Class 11

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2

Question 1.
Find the distance between the following pairs of points :
(i) (2, 3, 5) and (4, 3, 1)
(ii) (- 3, 7, 2) and (2, 4, – 1)
(iii) (- 1, 3, – 4) and (1, – 3, 4)
(iv) (2, – 1, 3) and (- 2, 1, 3).
Answer.
The distance between point P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is given by
PQ = \(\)
(i) Distance between points (2, 3, 5) and (4, 3, 1)
= \(\sqrt{(4-2)^{2}+(3-3)^{2}+(1-5)^{2}}=\sqrt{(2)^{2}+(0)^{2}+(-4)^{2}}\)
= \(\sqrt{4+16}=\sqrt{20}=2 \sqrt{5}\)

(ii) Distance between points (- 3, 7, 2) and (2, 4, – 1)
= \(\sqrt{(2+3)^{2}+(4-7)^{2}+(-1-2)^{2}}=\sqrt{(5)^{2}+(-3)^{2}+(-3)^{2}}\)
= \(\sqrt{25+9+9}=\sqrt{43}\)

(iii) Distance between points ( – 1, 3, – 4) and (1, – 3, 4)
= \(\sqrt{(1+1)^{2}+(-3-3)^{2}+(4+4)^{2}}\)
= \(\sqrt{(2)^{2}+(-6)^{2}+(8)^{2}}\)
= \(\sqrt{4+36+64}=\sqrt{104}=2 \sqrt{26}\)

(iv) Distance between points (2, -1, 3) and (-2, 1, 3)
= \(\sqrt{(-2-2)^{2}+(1+1)^{2}+(3-3)^{2}}\)
= \(\sqrt{(-4)^{2}+(2)^{2}+(0)^{2}}\)
= \(\sqrt{16+4}=\sqrt{20}=2 \sqrt{5}\)

Question 2.
Show that the point (- 2, 3, 5), (1, 2, 3) and (7,0, – 1) are collinear.
Answer.
Let points (- 2, 3, 5), (1, 2, 3) and (7, 0, – 1) be denoted by P, Q, and R respectively.
Points P, Q, and R are collinear if they lie on a line.

Here, PQ + QP = \(\sqrt{14}+2 \sqrt{14}=3 \sqrt{14}\) = PR
Hence, points P (- 2, 3, 5), Q(1, 2, 3), and P (7, 0, – 1) are collinear.

Question 3.
Verify the following:
(i) (0, 7, – 10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (- 1, 6, 6) and (- 4, 9, 6) are the vertices of a right angled triangle.
(iii) (- 1, 2, 1), (1, – 2, 5), (4, – 7, 8) and (2, – 3, 4) are the vertices of a parallelogram.
Answer.
(i) Let points (0, 7, – 10), (1, 6, – 6) and (4, 9, – 6) be denoted by A, B and C respectively.

Here, AB = BC ≠ CA.
Thus, the given points are the vertices of an isosceles triangle.

(ii) Let (0, 7, 10), (- 1, 6, 6) and (- 4, 9, 6) be denoted by A, B, and C respectively.

Now, AB² + BC² = (3√2)² + (3√2)²
= 18 + 18 = 36 = AC²
Hence, the given points are the vertices of a right-angled triangle.

(iii) Let points (- 1, 2, 1), (1, – 2, 5), (4, – 7, 8) and (2, – 3, 4) be denoted by A, B, C and D respectively.

Now, AB = CD = 6, BC = AD = √43.
Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.
Therefore, ABCD is a parallelogram.
Hence, the given points are the vertices of a parallelogram.

Question 4.
Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, – 1).
Answer.
Let P(x, y, z) be the point that is equidistant from points A(l, 2, 3) and B(3, 2, -1).
Accordingly, PA = PB
⇒ PA² = PB²
⇒ (x – 1)² + (y – 2)² + (z – 3)² = (x – 3)² + (y – 2)² + (z + 1)²
x² – 2x + 1 + y² – 4y + 4 + z² – 6z + 9 = x² – 6x + 9 + y² – 4y + 4 + z² + 2z +1
⇒ – 2x – 4y – 6z + 14 = – 6x – 4y + 2z + 14
⇒ – 2x – 6z + 6x – 2z = 0
⇒ 4x – 8z = 0
⇒ x – 2z = 0
Thus, the required equation is x – 2z = 0.

Question 5.
Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (- 4, 0, 0) is equal to 10.
Answer.
We have, a point P(x, y, z) such that, PA + PB = 0

On squaring both sides, we get
x² + y² + z² – 8x + 16 = 100 + x² + y² + z² + 8x + 16 – 20 \(\sqrt{x^{2}+y^{2}+z^{2}+8 x+16}\)
⇒ – 16x – 100 = – 20 \(\sqrt{x^{2}+y^{2}+z^{2}+8 x+16}\)
⇒ 4x + 25 = 5 \(\sqrt{x^{2}+y^{2}+z^{2}+8 x+16}\) [dividing both sides by -4]
Again squaring on both sides, we get
16x² + 200x + 625 = 25 (x² + y² + z² + 8x +16)
⇒ 16x² + 200x + 625 = 25x² + 25y² + 25z² +200x + 400
⇒ 9x² + 25y² + 25z² – 225 = 0.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.1

Question 1.
A point is on the x-axis. What are its y – coordinates and z – coordinates?
Answer.
Coordinates of any point on the X-axis is (x, 0, 0).
Because at X-axis, both y and z – coordinates are zero.
So, its y and z – coordinates are zero.

Question 2.
A point is in the XZ – plane. What can you say about its y-coordinate?
Answer.
Any point on the XZ – plane will have the coordinate (x, 0, z), so its y – coordinate is 0.

Question 3.
Name the octants in which the following points lie : (1, 2, 3), (4, – 2, 3), (4, – 2, – 5), (4, 2, – 5), (- 4, 2, – 5), (- 4, 2, 5), (- 3, – 1, 6), (2, – 4, – 7).
Answer.
The x-coordinate, y-coordinate, and z-coordinate of point (1, 2, 3) are all positive. Therefore, this point lies in octant I.
The x-coordinate, y-coordinate, and z-coordinate of point (4, – 2, 3) are positive, negative, and positive respectively. Therefore, this point lies in octant IV.
The x-coordinate, y-coordinate, and z-coordinate of point (4, – 2, – 5) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.
The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, – 5) are positive, positive, and negative respectively. Therefore, this point lies in octant V.
The x-coordinate, y-coordinate, and z-coordinate of point (- 4, 2, – 5) are negative, positive, and negative respectively. Therefore, this point lies in octant VI.
The x-coordinate, y-coordinate, and z-coordinate of point (- 4, 2, 5) are negative, positive, and positive respectively. Therefore, this point lies in octant II.
The x-coordinate, y-coordinate, and z-coordinate of point ( – 3, – 1, 6) are negative, negative, and positive respectively. Therefore, this point lies in octant III.
The x-coordinate, y-coordinate, and z-coordinate of point (2, – 4, – 7) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

Question 4.
Fill in the blanks :
(i) The x-axis and y-axis taken together determine a plane known as
(ii) The coordinates of points in the XY-plane are of the form
(iii) Coordinate planes divide the space into octants.
Answer.
(i) XY plane
(ii) (x, y, 0)
(iii) eight

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 11 Conic Sections Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 11 Conic Sections Miscellaneous Exercise

Question 1.
If a parabolic refletor is 20 cm in diameter and 5 cm deep, find the focus.
Answer.

Taking vertex of the parabolic reflector at origin, X-axis along the axis of parabola.
The equation of the parabola is of the fonn y² = 4ax.
Given, depth is 5 cm and diameter is 20 cm.
∵ Point P(5, 10) lies on parabola.
∴ (10)² = 4a(5) = a = 5
Clearly, focus is at the mid-point of given diameter i.e., S(5, 0).

Question 2.
An arch is in the form of a parabola with its Rxi vertical. The arch is 10 m high and 5 m wide at the base. How wide Is it 2m from the vertex of the parabola?
Answer.

Let the vertex of the parabola be at the origin and axis be along OY.
Then, the equation of the parabola is
x² = 4ay
The coordinates of end A of the (2.5, 10) and it lies on the eq. (i).
∴ (25)² = 4a × 10
⇒ a = \(\frac{6.25}{40}=\frac{5}{32}\) ……………(ii)

On putting the value of a from eq. (ii) in eq.(i), we get
x² = 4(\(\frac{5}{32}\))y ………………..(iii)

On substituting y = 2 in eq. (iii), we get
x² = \(\frac{5}{8}\) × 2
⇒ x² = \(\frac{5}{4}\)
⇒ x = \(\frac{\sqrt{5}}{2}\) m.
Hence, the width of the arc at a height of 2 m from vertex is 2 × \(\frac{\sqrt{5}}{2}\) i.e., √5 m.

Question 3.
The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6m. Find the length of a supporting wire attached to the roadway 18 m from the middle.
Answer.

The cable is in the form of a parabola x² = 4ay.
Focus is at the middle of the cable, the shortest and longest vertical supports are 6 m and 30 m and roadway is 100 m long.
Since, point A(50, 24) lies on parabola x² = 4ay.
(50)² = 4a(24)
= 4a = 625
Equation of parabola is x² = \(\frac{625}{6}\) y
[Put 4a = \(\frac{625}{6}\)]
Let the support at 18 m from middle be l m, then B(18, l – 6) lies on the parabola.
∴ (18)² = \(\frac{625}{6}\) (l – 6)
l = \(\frac{18 \times 18 \times 6}{625}\) + 6
= 3.11 + 6 = 9.11 (approx)
Hence, the length of supporting wire is 9.11 m.

Question 4.
An arch is in the form of a semi-ellipse. It is 8 m wide and 2m high at the centre. Find the height of the arch at a point 1.5 m from one end.
Answer.

Clearly, equation of ellipse is of the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 ……………(i)
Here, it is given that, 2a = 8 and b = 2.
⇒ a = 4 and b =2
On putting the values of a and b in eq. (i), we get
\(\frac{x^{2}}{16}+\frac{y^{2}}{4}\) = 1
which is required equation.
Given, AP = 1.5 m
∴ OP = OA – AP = 4 – 15
OP = 25m
Let PQ = k
∴ Coordinate of Q (2.5, k) will satisfy the equation of ellipse.

Question 5.
A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.
Answer.

Let l be the length of the rod and at any position meet X-axis at A(a, 0) and Y-axis at B(0, b), so that
l² = a² + b²
⇒ (12)² = a² + b² …………….(i) [∵ l = 12]
Let P be the point on AB which is 3 cm from A and hence 9 cm from B.

This means that P divides AB in ratio 3 : 9 i.e., 1 : 3.
If P = (x, y), then by section formula, we have

(x, y) = \(\left(\frac{1 \times 0+3 \times a}{1+3}, \frac{1 \times b+3 \times 0}{1+3}\right)\)

(x, y) = \(\left(\frac{3 a}{4}, \frac{b}{4}\right)\)

⇒ x = \(\frac{3 a}{4}\)
⇒ a = \(\frac{4 x}{3}\) and b = 4y
On putting the values of a and b in eq. (i), we get
144 = (\(\frac{4 x}{3}\))² + (4 y)²
⇒ 1 = \(\frac{x^{2}}{9 \times 9}+\frac{y^{2}}{9}\)

⇒ \(\frac{x^{2}}{81}+\frac{y^{2}}{9}\) = 1
which is required equation.

Question 6.
Find the area of the triangle formed by the lines joining the vertex of the parabola x² = 12y to the ends of its latus rectum.
Answer.
The given parabola is x² = 12y.
On comparing this equation with x² = 4ay, we obtain
4a = 12
⇒ a = 3.
The coordinates of foci are S (0, a) = S (0, 3)

Let AB be the latus rectum of the given parabola.
The given parabola can be roughly drawn as
At y = 3, x² = 12 (3)
x² = 36
⇒ x = ± 6.
The coordinates of A are (- 6, 3), while the coordinates of B are (6, 3).
Therefore, the vertices of ∆OAB are 0 (0, 0), A (- 6, 3), and B (6, 3).
Area of ∆OAB = \(\frac{1}{2}\) |0 (3 – 3) + (- 6) (3 – 0) + 6 (0 – 3)|
= \(\frac{1}{2}\) |(- 6)(3) + 6(- 3)| unit²
= \(\frac{1}{2}\) |- 18 – 18| unit²
= \(\frac{1}{2}\) |- 36| unit²
= \(\frac{1}{2}\) × 36 unit²
= 18 unit²
Thus, the required area of the triangle is 18 unit².

Question 7.
A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.
Answer.

Let A and B be the positions of the two flag posts and P(x, y) be the position of the man. Accordingly, PA + PB = 10.
We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis is 10 m, while points A and B are the foci.
Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the x-axis, the ellipse can be diagrammatically represented as.
The equation of the ellipse will be of the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 , where a is the semi-major axis.
Accordingly, 2a = 10
⇒ a = 5.
Distance between the foci, 2c = 8
⇒ c = 4
On using the relation c = \(\sqrt{a^{2}-b^{2}}\) we obtain 4 = \(\sqrt{25-b^{2}}\)
⇒ 16 = 25 – b²
⇒ b = 25 – 16 = 9
⇒ b = 3
Thus, the equation of the path traced by the man is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1.

Question 8.
An equilateral triangle is inscribed in the parabola y =4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
Answer.
Given, equation of parabola is y² = 4ax.
Let p be the side of equilateral ∆OAB, whose one vertex is the vertex of parabola.
Then, by symmetry, AB is perpendicular to the axis ON of parabola.
Let ON = x, then BN = \(\frac{p}{2}\)
Since, B(x, \(\frac{p}{2}\)) lies on parabola.

⇒ p² = 64 × 3 × a²
⇒ p = 8a√3
Hence, side of an equilateral triangle is 8a√3.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 11 Conic Sections Ex 11.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4

Question 1.
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola
\(\frac{x^{2}}{16}-\frac{y^{2}}{9}\) = 1.
Answer.
The given equation is \(\frac{x^{2}}{16}-\frac{y^{2}}{9}\) = 1 or \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.

On comparing this equation with the standard equation of hyperbola i.e., \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, we obtain a = 4 and b = 3.

We know that
a² + b² = c²
4² + 3² = 25
⇒ c = 5
Therefore, The coordinates of the foci are (± 5, 0).
The coordinates of the vertices are (± 4, 0).
Eccentricity e = \(\frac{c}{a}=\frac{5}{4}\)

Length of latusrectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 9}{4}=\frac{9}{2}\).

Question 2.
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola
\(\frac{y^{2}}{9}-\frac{x^{2}}{27}\) = 1.
Answer.
The given equation is \(\frac{y^{2}}{9}-\frac{x^{2}}{27}\) = 1 or \(\frac{y^{2}}{3^{2}}-\frac{x^{2}}{(\sqrt{27})^{2}}\) = 1.
On comparing this equation with the standard equation of hyperbola i.e., \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1, we obtain a = 3 and b = √27.

We know that a² + b² = c²
c² = 3² + (√27)²
= 9 + 27 = 36
⇒ c = 6
Therefore, the coordinates of the foci are (0, ± 6).
The coordinates of the vertices are (0, ± 3).
Eccentricity, e = \(\frac{c}{a}=\frac{6}{3}\) = 2.
Length of latusrectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 27}{3}\) = 18.

Question 3.
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola
9y² – 4x² = 36.
Answer.
The given equation is 9y² – 4x² = 36
It can be written as 9y² – 4x² = 36
or \(\frac{y^{2}}{4}-\frac{x^{2}}{9}\) = 1

or \(\frac{y^{2}}{2^{2}}-\frac{x^{2}}{3^{2}}\) = 1 ………….(i)
On comparing equation (i) with the standard equation of hyperbola i.e., \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1, we obtain a = 2 and b = 3.

We know that a² + b² = c²
c² = 4 + 9 = 13
⇒ c = √13.
Therefore, the coordinates of the foci are (0, ± √13).
The coordinates of the vertices are (0, ± 2).
Eccentricity, e = \(\frac{c}{a}=\frac{\sqrt{13}}{2}\)

Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 9}{2}\) = 9.

Question 4.
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 16x² – 9y² = 576.
Answer.
The given equation is 16x² – 9y² = 576
It can be written as 16x² – 9y² = 576

⇒ \(\frac{x^{2}}{36}-\frac{y^{2}}{64}\) = 1

⇒ \(\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}}\) = 1 ………………(i)

On comparing equation (i) with the standard equation of hyperbola i.e.,
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, we obtain a = 6 and b = 8.
We know that a² + b² = c²
c² = 36 + 64 = 100
⇒ c = 10
Therefore, the coordinates of the foci are (± 10, 0).
The coordinates of the vertices are (± 6, 0).
Eccentricity, e = \(\frac{c}{a}=\frac{10}{6}=\frac{5}{3}\).

Length of larus rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 64}{6}=\frac{64}{3}\).

Question 5.
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5y² – 9x² = 36.
Answer.
The given equation is 5y² – 9x² = 36.
⇒ \(\frac{y^{2}}{\left(\frac{36}{5}\right)}-\frac{x^{2}}{4}\) = 1

⇒ \(\frac{y^{2}}{\left(\frac{6}{\sqrt{5}}\right)^{2}}-\frac{x^{2}}{2^{2}}\) = 1
On comparing equation (i) with the standard equation of hyperbola i.e., \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1,

we obtain a = \(\frac{6}{\sqrt{5}}\) = and b = 2.

We know that a² + b² = c².
∴ c² = \(\frac{36}{5}+4=\frac{56}{5}\)

⇒ c = \(\sqrt{\frac{56}{5}}=\frac{2 \sqrt{14}}{\sqrt{5}}\)

Therefore, the coordinates of the foci are (0, ± \(\frac{2 \sqrt{14}}{\sqrt{5}}\))

The coordinates of the vertices are (o, ± \(\frac{6}{\sqrt{5}}\)).

Eccentricity, e = \(\frac{c}{a}=\frac{\left(\frac{2 \sqrt{14}}{\sqrt{5}}\right)}{\left(\frac{6}{\sqrt{5}}\right)}=\frac{\sqrt{14}}{3}\)

Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 4}{\left(\frac{6}{\sqrt{5}}\right)}=\frac{4 \sqrt{5}}{3}\)

Question 6.
Find the coordinates of the foci, and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola
49y² – 16x² = 784.
Answer.
The given equation is 49y² – 16x² = 784
it can be writtern as 49y² – 16x² = 784

or \(\frac{y^{2}}{16}-\frac{x^{2}}{49}\) = 1

or \(\frac{y^{2}}{4^{2}}-\frac{x^{2}}{7^{2}}\) = 1 ……………….(i)

On comparing equation (i) with the standard equation of hyperbola i.e., \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1,
we obtain a = 4 and b = 7.

We knowthat, a² + b² = c².
∴ c² = 16 + 49 = 65
⇒ c = √65
Therefore, The coordinates of the foci are (0, ± √65
The coordinates of the vertices are (0, ± 4).

Eccentricity, e = \(\frac{c}{a}=\frac{\sqrt{65}}{4}\).

Length of lattis rectum = \(\frac{2 b^{2}}{a}=\frac{2 \times 49}{4}=\frac{49}{2}\).

Question 7.
Find the equation of the hyperbola satisfying the given conditions : Vertices (± 2, 0), foci (± 3, 0).
Answer.
Given, vertices (± 2, 0), foci (± 3, 0)
Here, the vertices are on the x-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
Since the vertices are (± 2, 0), a = 2
Since the foci are (± 3, 0), c = 3
We know that a² + b² = c².
∴ 2² + b² = 3²
b² = 9 – 4 = 5
Thus, the equation of the hyperbola is \(\frac{x^{2}}{4}-\frac{y^{2}}{5}\)= 1.

Question 8.
Find the equation of the hyperbola satisfying the given conditions : Vertices (0, ± 5), foci (0, ± 8).
Ans.
We have,vertices (0, ± a) = (0, ± 5) = a = 5
and foci(0, ± c) = (0, ± 8)
⇒ c = 8
We know that, c² = a² + b²
⇒ 64 = 25 + b²
⇒ b² = 64 – 25
⇒ b² = 39
Here, the foci and vertices lie on Y-axis, therefore the equation of hyperbola is of the form \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1
i.e. \(\frac{y^{2}}{25}-\frac{x^{2}}{39}\) = 1

Question 9.
Find the equation of the hyperbola satisfying the given conditions : Vertices (0, ± 3), foci (0, ± 5).
Answer.
Given, vertices (0, ± 3), foci (0, ± 5)
Here, the vertices are on the y-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1.
Since the vertices are (0, ± 3), a = 3.
Since the foci are (0, ± 5), c = 5.
We know that a² + b² = c².
3² + b² = 5²
⇒ b² = 25 – 9 = 16.
Thus, the equation of the hyperbola is \(\frac{y^{2}}{9}-\frac{x^{2}}{16}\) = 1.

Question 10.
Find the equation of the hyperbola satisfying the given conditions : Foci (± 5, 0), the transverse axis is of length 8.
Answer.
Given, foci (± 5, 0), the transverse axis is of length 8.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
Since the foci are (± 5, 0), c = 5.
Since the length of the transverse axis is 8, 2a = 8
⇒ a = 4.
We know that, a² + b² = c².
∴ 4² + b² = 5²
⇒ b² = 25 – 16 = 9

Question 11.
Find the equation of the hyperbola satisfying the given conditions : Foci (0, ± 13), the conjugate axis is of length 24. Answer.
Given, foci (0, ± 13), the conjugate axis is of length 24.
Here, the foci are on the y-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}\) = 1.
a b2
Since the foci are (0, ± 13), c = 13.
Since the length of the conjugate axis is 24, 2b = 24
⇒ b = 12.
We know that a² + b² = c².
a² + 12² = 13²
⇒ a² = 169 – 144 = 25
Thus, the equation of the hyperbola is \( \frac{y^{2}}{25}-\frac{x^{2}}{144}\) = 1.

Question 12.
Find the equation of the hyperbola satisfying the given conditions : Foci (± 3√5, 0), the latus rectum is of length 8.
Answer.

Here, foci at (± 3√5, 0)
∴ (± c, 0) = (± 3√5, 0)
⇒ c = 3√5
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = 8 [given]
∴ b² = 4a …………….(i)
We know that, c² = a² + b²
⇒ (3√5)² = a² + 4a [put c = 3√5]
⇒ a = – 9, a = 5
∴ a = 5 as a cannot be negative.
On putting a = 5 in eq. (i), we get
b² = 5 × 4
⇒ b² = 20
Since, foci lies on X-axis, therefore the equation of hyperbola is of the form
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1

i.e., \(\frac{x^{2}}{25}-\frac{y^{2}}{20}\) = 1 [Put a² = 5 and b² = 20]
⇒ 20x² – 25y² = 500
⇒ 4x² – 5y² = 100 [dividing both sides by 5]
which is required equation of hyperbola.

Question 13.
Find the equation of the hyperbola satisfying the given conditions : Foci (± 4, 0), the latus rectum is of length 12.
Answer.
Given, foci (± 4, 0), the latus rectum axis is of length 12.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
Since the foci are (± 4, 0), c = 4.
Length of latus rectum =12
⇒ \(\frac{2 b^{2}}{a}\) = 12
⇒ b² = 6a
We know that a² + b² = c².
a² + 6a = 16
⇒ a² + 6a – 16 = 0
⇒ a² + 8a – 2a – 16 = 0
⇒ (a + 8) (a – 2) = 0
⇒ a = – 8, 2
Since a is non-negative, a = 2.
b² = 6a = 6 × 2 = 12
Thus, the equation of the hyperbola is \(\frac{x^{2}}{4}-\frac{y^{2}}{12}\) = 1.

Question 14.
Find the equation of the hyperbola satisfying the given conditions : Vertices (± 7, 0), e = \(\frac{4}{3}\).
Answer.
Given, vertices (± 7, 0), e = \(\frac{4}{3}\).
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.
Since the vertices are (± 7, 0), a = 7.

Question 15.
Find the equation of the hyperbola satisfying the given conditions : Foci (0, ± √10), passing through (2, 3).
Answer.
Since, the foci (0, ± √10) are along Y-axis, then equation of hyperbola is of the form

9(10 – a²) – 4a² = a²(10 – a²)
⇒ 90 – 13a² = 10a² – a⁴
⇒ a⁴ – 23a² + 90 = a
⇒ (a² – 18) (a² – 5) = 0
∴ a² = 18 or 5
Since, e = \(\frac{\sqrt{10}}{a}\) and e > 1
∴ \(\frac{\sqrt{10}}{a}\) > 1
⇒ a < √10; a² < 10
we reject a² = 18
Thus, a² = 5
∴ b² = a² e² – a²
= 10 – 5 = 5
Hence, equation of hyperbola is \(\frac{y^{2}}{5}-\frac{x^{2}}{5}\) = 1 [put a² = b² = 5]
or y² – x² = 5.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 11 Conic Sections Ex 11.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2

Question 1.
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length .of the latus rectum for y² = 12x.
Answer.
The given equation is y² = 12x.
Here, the coefficient of x is positive.
Hence, the parabola opens towards the right.
On comparing this equation with y² = 4 ax, we obtain
4a = 12
⇒ a = 3.
∴ Coordinates of the focus = (a, 0) = (3, 0)
Since the given equation involves y², the axis of the parabola is the x-axis.
Equation of directrix, x = – a i.e., x = – 3 i.e., x + 3 = 0
Length of latus rectum = 4a = 4 × 3 = 12.

Question 2.
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for x² = 6y.
Answer.
The given equation is x² = 6y.
Here, the coefficient of y is positive. Hence, the parabola opens upwards.
On comparing this equation with x² = 4ay , we obtain
4a = 6
⇒ a = \(\frac{3}{2}\).
Coordinates of the focus = (0, a) = (0, \(\frac{3}{2}\))
Since the given equation involves x , the axis of the parabola is the y-axis.
Equation of directrix, y = – a i.e., y = – \(\frac{3}{2}\)
Length of latus rectum = 4a = 6.

Question 3.
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for y² = – 8x.
Answer.
Here, the coefficient of x is negative.
Hence, the parabola opens towards the left.
On comparing this equation with y² = – 4 ax , we obtain
– 4a = – 8
⇒ a = 2
Coordinates of the focus = (- a, 0) = (- 2, 0)
Since the given equation involves y², the axis of the parabola is the x-axis.
Equation of directrix, x = a i.e., x = 2
∴ Length of latus rectum = 4a = 8.

Question 4.
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for x² = – 16y.
Ans.
The given equation is x² = – 16y
Here, the coefficient of y is negative.
Hence, the parabola opens downwards.
On comparing this equation with x² = – 4ay, we obtain
– 4a = – 16
⇒ a = 4
∴ Coordinates of the focus = (0, – a) = (0, – 4)
Since the given equation involves x², the axis of the parabola is the y-axis.
Equation cf directrix, y = a i.e., y = 4
Length of latus rectum = 4a = 16.

Question 5.
Find the coordinntes of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for y² = 10x.
Answer.
The given equation is y² = 10x.
Here, the coefficient of x is positive.
Hence, the parabola opens towards the right.
On comparing this equation with y² = 4ax, we obtain
4a = 10
⇒ a = \(\frac{5}{2}\)
Coordinates of the focus = (a, 0) = (\(\frac{5}{2}\), 0)
Since the given equation involves y², the axis of the parabola is the x-axis.
Equation of directrix, x = – a, i.e., x = – \(\frac{5}{2}\)
Length of latus rectum = 4a = 10.

Question 6.
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for x\(\frac{5}{2}\) = – 9y.
Answer.
We have, equation of parabola is x² = – 9y
So, axis of symmetry is along x-axis.
The coefficient of y is negative, so parabola opens downward.
On comparing with the equation,
x² = – 4ay, we get a = \(\frac{9}{4}\)
Thus, the focus is (0, – a) i.e., (0, – \(\frac{9}{4}\))
Equation of directrix is y = a i.e., y = 9/4.
Length of latus rectum = 4a i.e., 9.

Question 7.
Find the equation of the parabola that satisfies the given conditions : Focus (6, 0): directrix x = – 6.
Answer.
Given, focus (6, 0) and directrix, x = – 6.
Here, we see that in focus, x-coordinate is positive and y-coordinate is zero.
So, focus lies on the positive direction of X-axis i.e., equation of parabola will be of the form y² = 4ax with a = 6.
Hence, required equation is
y² = 4 × 6x
⇒ y² = 24x

Question 8.
Find the equation of the parabola that satisfies the given conditions : Focus (0, – 3); directrix y = 3.
Answer.
Given, focus = (0, – 3); directrix y=3.
Since the focus lies on the y-axis, the y-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form x² = 4 ay or x² = – 4ay.
It is also seen that the directrix, y = 3 is above the x-axis, while the focus (0, – 3) is below the x-axis.
Hence, the parabola is of the form x²= – 4ay.
Here, a = 3.
Thus, the equation of the parabola is x² = – 12y.

Question 9.
Find the equation of the parabola that satisfies the given conditions : Vertex (0, 0); focus (3, 0).
Answer.
Given, vertex (0, 0); focus (3, 0).
Since the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y² = 4ax.
Since the focus is (3, 0), a = 3.
Thus, the equation of the parabola is y² = 4 × 3 × x, i.e., y² = 12x.

Question 10.
Find the equation of the parabola that satisfies the given conditions: Vertex (0, 0) focus (-2, 0).
Answer.
Given, vertex (0, 0); focus (- 2, 0)
Since the vertex of the parabola is (0, 0) and the focus lies on the negative x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y² = – 4ax.
Since the focus is (- 2, 0), a = 2.
Thus, the equation of the parabola is y² = – 4(2)x, i.e., y = – 8x.

Question 11.
Find the equation of the parabola that satisfies the given conditions: Vertex (0, 0) passing through (2, 3) and axis is along x-axis.
Answer.
Since the vertex is (0, 0) and the axis of the parabola is the x-axis, the equation of the parabola is either of the form
y² = 4ax or y² = – 4ax.
The parabola passes through point (2, 3), which lies in the first quadrant.
Therefore, the equation of the parabola is of the form y² = 4 ax, while point (2, 3) must satisfy the equation y² = 4ax.
3² = 4a(2)
a = \(\frac{9}{8}\)
Thus, the equation of the paraola is y² = – 4(2)x, i.e., y² = – 8x.

Question 12.
Find the equation of the parabola that statisfies the given conditions; Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.
Answer.

Since, the parabola is symmetric about Y-axis and has its vertex at the origin, so the equation is of the form x² = 4ay or x² = – 4ay.
But parabola passes through (5, 2) which lies in the first quadrant. It must open upward.
Thus, the equation is of the form
x² = 4ay ……………….(i)
Since, parabola passes through (5, 2)
(5)² = 4a(2)
⇒ 25 = 8a
⇒ a = \(\frac{25}{8}\)
On putting the value of a in eq.(i), we get
x² = 4 (\(\frac{25}{8}\)) y
⇒ x² = \(\frac{25}{8}\) y

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 11 Conic Sections Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1

Question 1.
Find the equation of the circle with centre (0, 2) and radius 2.
Answer.
The equation of a circle with centre (h, k) and radius r is given as
(x – h)² + (y – k)² = r²
It is given that centre (h, k) = (0, 2) and radius (r) = 2
Therefore, the equation of the circle is
(x – 0)² + (y – 2)² = 2²
x² + y² + 4 – 4y = 4
x² + y² – 4y = 0.

Question 2.
Find the equation of the circle with centre (- 2, 3) and radius 4.
Answer.
The equation of a circle with centre (h, k) and radius r is given as
(x – h)² + (y – k)² = r²
It is given that centre (h, k) = (- 2, 3) and radius (r) = 4.
Therefore, the equation of circle is
(x + 2)² + (y – 3)² = (4)²
x² + 4x + 4 + y² – 6y + 9 = 16
x² + y² + 4x – 6y – 3 = 0.

Question 3.
Find the equation of the circle with centre (\(\frac{1}{2}\), \(\frac{1}{4}\)) and radius \(\frac{1}{12}\).
Answer.

Question 4.
Find the equation of the circle with cehtre (1,1) and radius √2.
Answer.

Here, centre is at (1, 1).
∴ h = 1, k = 1
Also, radius, r = √2 units
[∵ equation of circle having centre (h, k) and radius r is (x – h)² + (y – k)² = r²].
Equation of circle is (x – 1)² + (y – 1)² = (V2)²
⇒ x² – 2x + 1 + y² – 2y + 1 = 2
⇒ x² + y² – 2x – 2y = 0

Question 5.
Find the equation of the circle with centre (- a, – b) and radius \(\sqrt{a^{2}-b^{2}}\).
Answer.
The equation of a circle with centre (h, k) and radius r is given as (x – h)² + (y – k)² = r²
It is given that centre (h, k) = (- a, – b) and radius (r) = \(\sqrt{a^{2}-b^{2}}\).
Therefore, the equation of the circle is (x + a)² + (y + b)² = (\(\sqrt{a^{2}-b^{2}}\))²
x² + 2ax + a² + y² + 2by + b² = a² – b²
x² + y² + 2ax + 2by + 2b² = 0.

Question 6.
Find the centre and radius of the circle (x + 5)² + (y – 3)² = 36.
Answer.
The equation of given circle is (x + 5)² + (y – 3)² = 36
(x + 5)² + (y – 3)² = 36
⇒ {x – (- 5)}² + (y – 3)² = (6)²,
which is of the form (x – h)² + (y – k)² = r²,
where h = – 5, k = 3, and r = 6.
Thus, the centre of the given circle is (- 5, 3), while its radius is 6.

Question 7.
Find the equation of the circle x² + y² – 4x – 8y – 45 = 0.
Answer.
The equation of given circle is x² + y² – 4x – 8y – 45 = 0.
x² + y² – 4x – 8y – 45 = 0
⇒ (x² – 4x) + (y² – 8y) = 45
⇒ {x² – 2 (x) (2) + 22} + {y² – 2(y) (4) + 42} – 4 – 16 = 45
⇒ (x – 2)² + (y – 4)² = 65 ,
⇒ (x – 2)² + (y – 4)² = (√65)², which is of the form
(x – h)² + (y – k)² = r² , where h = 2, k = 4 and r = √65
Thus, the centre of the given circle is (2, 4), while its radius is √65.

Question 8.
Find the centre and radius of the circle x² + y² – 8x + 10y – 12 = 0.
Answer.
The equation of the given circle is x² + y² + 8x + 10y – 12 = 0.
x² + y² – 8x + 10y – 12 = 0
(x² – 8x) + (y² + 10y) = 12
⇒ (x² – 2 (x) (4) + 4²) + (y² + 2(y) (5) + 5²) – 16 – 25 = 12
(x – 4)² + (y + 5)² = 53
(x – 4)² + {y – (- 5)}² = (√53)², which is of the form
(x – h)² + (y – k)² = r², where h = 4, k = – 5 and r = √53
Thus, the centre of the given circle is (2, 4), while its radius is √53.

Question 9.
Find the centre and radius of the circle 2x² + 2y² – x = 0.
Answer.
The given equation of circle is
2x² + 2y² – x = 0
x² + y² – \(\frac{x}{2}\) = 0
(x² – \(\frac{x}{2}\)) + y² = 0
On adding \(\frac{1}{16}\) to make perfect squares, we get
(x² – \(\frac{x}{2}\) + \(\frac{1}{16}\)) + y² = \(\frac{1}{16}\)
(x – \(\frac{1}{4}\))² + (y – 0)² = (\(\frac{1}{4}\))²
On comparing with (x – h)² + (y – k)² = r² , we get
h = \(\frac{1}{4}\), k = 0 and r = \(\frac{1}{2}\)
Centre = (h, k) = (\(\frac{1}{4}\), 0)
Radius = \(\frac{1}{4}\)

Question 10.
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Answer.
Let the equation of the required circle be (x – h)² + (y – k)² = r².
Since the circle passes through points (4, 1) and (6, 5).
(4 – h)² + (1 – k)² = r² ……………(i)
(6 – h)² + (5 – k)² = r² ……………(ii)
Since the centre (h, k) of the circle lies on line 4x + y = 16,
4h + k = 16
From equations (i) and (ii), we obtain
(4 – h)² + (1 – k)² = (6 – h)² + (5 – k)²
⇒ 16 – 8h + h² + 1 – 2k + k² = 36 – 12h + h² + 25 – 10k + k²
⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k
⇒ 4h + 8k = 44
⇒ h + 2k = 11 …………………(iv)
On solving equations (iii) and (iv), we obtain h = 3 and k = 4.
On substituting the values of h and k in equation (i), we obtain
(4 – 3)² + (1 – 4)² = r²
⇒ (1)² + (- 3)² = r²
⇒ 1 + 9 = r²
⇒ r² = 10
⇒ r = √10
Thus, the equation of the required circle is (x – 3)² + (y – 4)²= (√10)²
x² – 6x + 9 + y² – 8y + 16 = 10
x² + y² – 6x – 8y + 15 = 0.

Question 11.
Find the equation of the circle passing through the points (2, 3) and (- 1, 1) and whose centre is on the line x – 3y – 11 = 0.
Answer.

Let the equation of the circle be
(x -h)² + (y – k)² =r² ……………(i)
Since, the circle (i) passes through (2, 3) and (- 1, 1).
We have, (2 – h)² + (3 – k)² = r²
⇒ h² – 4h + 4 + 9 – 6k + k² = r² and
(- 1 – h)² + (1 – k)² = r²
h² + 2h + 1 + 1 – 2k + k² = r²
Centre (h, k) of circle is on the line
x – 3y – 11 = 0
∴ h – 3k – 11 = 0 …………..(iv)
On subtracting eQuestion (iii) from eq.(ii), we get
– 6h – 4k + 11 = 0
⇒ 6h + 4k = 11 ……………….(v)
On solving eqs. (iv) and (v), we get
h = \(\frac{7}{2}\), k = \(\frac{-5}{2}\)
On putting the values of h and k in eq. (ii), we get
(2 – \(\frac{7}{2}\))² + (3 + \(\frac{5}{2}\))² = r²

(- \(\frac{7}{2}\))² + (\(\frac{11}{2}\))² = r²

Therefore, equation of circle having centre (\(\frac{7}{2}\), – \(\frac{5}{2}\)) and radius \(\frac{65}{2}\) is

(x – \(\frac{7}{2}\))² + (y + \(\frac{5}{2}\))² = \(\frac{65}{2}\)

x² + \(\frac{49}{4}\) – 7x + y² + \(\frac{25}{4}\) + 5y = \(\frac{65}{2}\)

x² + y² – 7x + 5y – \(\frac{56}{4}\) = 0
x² + y² – 7x + 5y – 14 = 0

Alternative Method:

Let the equation of circle be
x² + y² + 2gx + 2fy + c = 0 ……………..(i)
Since, it passes through (2, 3).
∴ (2)² + (3)² + 2g(2) + 2f(3) + c = 0
4 + 9 + 4g + 6f + c = 0
4g + 6f + c = – 13 ……………….(ii)
Circle also passes through (- 1, 1).
∴ (- 1)² + (1)² + 2g(- 1) + 2f(1) + c = 0
⇒ 1 + 1 – 2g + 2f + c = 0
⇒ – 2g + 2f + c = – 2 …………..(iii)
Since, centre of circle lies on line x – 3y – 11 = 0
i.e.,C(- g, – f) lies on line.
∴ – g – 3 (- f) – 11 = 0
– g + 3f = 11 …………..(iv)
On subtracting eq. (iii) from eq. (ii), we get
6g + 4f = – 11 …………….(v)
On solving eqs. (iv) and (y) for g and f, we get
g = – \(\frac{7}{2}\), f = \(\frac{5}{2}\)
On putting the values of g and f in eq. (ii), we get
⇒ – 14 + 15 + c = – 13
⇒ c = – 14
Now, putting the values of g, f and c in eq. (i), we get
x² + y² + 2(- \(\frac{7}{2}\)) + 2 (\(\frac{7}{2}\)) – 14 = 0
x² + y² – 7x + 5y – 14 = 0.

Question 12.
Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Answer.
Let the equation of the required circle be (x – h)² + (y – k)² = r².
Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.
Now, the equation of the circle becomes (x – h)² + y² = 25.
It is given that the circle passes through point (2, 3).
(2 – h)² + 3² = 25
⇒ (2 – h)² = 25 – 9
⇒ (2 – h)² = 16
⇒ 2 – h = ± √l6 = ± 4
If 2 – h = 4, then h = – 2
If 2 – h = – 4, then h = 6
when h = – 2, the equation of the circle becomes
(x + 2)² + y² = 25
x² + 4x + 4 + y² = 25
x² + y² + 4x – 21 = 0.
When h = 6, the equation of the circle becomes
(x – 6)² + y² = 25
x² – 12x + 36 + y² = 25
x² + y² – 12x + 11 = 0.

Question 13.
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
Answer.
Let the equation of the required circle be (x- h)² + (y – k)² = r².
Since the circle passes through (0, 0).
(0 – h)² + (0 – k)² = r².
h² + k² = r²
The equation of the circle now becomes (x – h)² + (y – k)² = h² + k².
It is given that the circle makes intercepts a and b on the coordinate axes.
This means that the circle passes thróugh points (a, 0) and (0, b).
(a – h)² + (0 – k)² = h² + k²
(0 – h)² + (b – k)² = h² + k²
From equation (i), we obtain
a² – 2ah + h² + k² = h² + k²
⇒ a² – 2ah = 0
⇒ a(a – 2h) = 0
⇒ a = 0 or (a – 2h) = 0
However, a ≠ 0; hence, (a – 2 h) = 0
⇒ h = \(\frac{a}{2}\).
From equation (ii), we obtain
h² + b² – 2bk + k² = h² + k²
⇒ b² – 2bk = 0
⇒ b(b – 2k) = 0
⇒ b = 0 or (b – 2k) = 0
However, b ≠ 0; hence, (b – 2k) = 0
⇒ k = \(\frac{b}{2}\).
Thus, the equation of the required circle is .
\(\left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{b}{2}\right)^{2}=\left(\frac{a}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2}\)

⇒ \(\left(\frac{2 x-a}{2}\right)^{2}+\left(\frac{2 y-b}{2}\right)^{2}=\frac{a^{2}+b^{2}}{4}\)

⇒ 4x² – 4ax + a² + 4y² – 4by + b² = a² + b²
⇒ 4x² + 4y² – 4ax – 4by = 0
⇒ x² + y² – ax – by = 0.

Question 14.
Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
Answer.
The equation of circle is (x – h)² + (y – k)² = r² ……………..(i)
Since the circle passes through point (4, 5) and co-ordinates of centre are (2, 2)
∴ radius of circle = \(\sqrt{(4-2)^{2}+(5-2)^{2}}=\sqrt{4+9}=\sqrt{13}\)
Now the equation of required circle is
(x – 2)² + (y – 2)² = (√13)²
4x² + 4y² – 4ax – 4ay = 13
x² + y² – 4x – 4y – 5 = 0

Question 15.
Does the point (- 2.5, 3.5) lie inside, outside or on the circle x² + y² = 25?
Answer.
The equation of the given circle is x² + y² = 25
x² + y² = 25
⇒ (x – 0)² + (y – 0)² = 5², which is of the form(x – h)² + (y – k)² = r²,
where h = 0, k = 0 and r = 5.
centre = (0, 0) and radius = 5.
Distance between point (- 2.5, 3.5) and centre (0, 0)
= \(\sqrt{(-2.5-0)^{2}+(3.5-0)^{2}}=\sqrt{6.25+12.25}=\sqrt{18.5}\)
= 4.3 (approx.) < 5.
Since the distance between point (- 2.5, 3.5) and centre (0, 0) of the circle is less than the radius of the circle, point (- 2.5, 3.5) lies inside the circle.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 10 Straight Lines Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 10 Straight Lines Miscellaneous Exercise

Question 1.
Find the values of k for which the line (k – 3) x – (4 – k²)y + k² – 7k + 6 = 0 is
(a) Parallel to the x-axis
(b) Parallel to the y-axis
(c) Passing through the origin.
Answer.
The given equation of line is
(k – 3) x – (4 – k²)y + k² – 7k + 6 = 0
(a) If the given line is parallel to the x-axis, then
Slope of the given line = Slope of the x-axis The given line can be written as
(4 – k²)y = (k – 3)x + k² – 7k + 6 = 0
y =\(\frac{(k-3)}{\left(4-k^{2}\right)} x+\frac{k^{2}-7 k+6}{\left(4-k^{2}\right)}\), which is of the form y = mx + c.

Slope of the given line = \(\frac{(k-3)}{\left(4-k^{2}\right)}\)

Slope of the x-axis = 0
\(\frac{(k-3)}{\left(4-k^{2}\right)}\) = 0

⇒ k – 3 = 0
⇒ k = 3
Thus, if the given line is parallel to the x-axis, then the value of k is 3.

(b) If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be undefined.
The slope of the given line is \(\frac{(k-3)}{\left(4-k^{2}\right)}\)

Now, \(\frac{(k-3)}{\left(4-k^{2}\right)}\) is undefined at k² = 4

k² = 4
⇒ k = ± 2
Thus, if the given line is parallel to the y-axis, then the value of k is ± 2.

(c) If the given line is passing through the origin, then point (0, 0) satisfies the given equation of line.
(k – 3) (0) – (4 – k²) (0) + k² – 7k + 6 = 0
k² – 7k + 6 = 0
k² – 6k – k + 6 = 0
(k – 6) (k – 1) = 0 k = 1 or 6
Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.

Question 2.
Find the value of 6 and p, if the equation x cos θ + y sin θ = p is the normal form of the line √3x + y + 2 = 0.
Answer.
The equation of the given line is √3x + y + 2 = 0.
This equation can be reduced as √3x + y + 2 = 0
⇒ On dividing both sides by \(\frac{-\sqrt{3} x-y=2}{(-\sqrt{3})^{2}+(-1)^{2}}\) = 2, we obtain

\(-\frac{\sqrt{3}}{2} x-\frac{1}{2} y=\frac{2}{2}\)

⇒ \(\left(-\frac{\sqrt{3}}{2}\right) x+\left(-\frac{1}{2}\right) y\) = 1

On comparing equation (i) to x cos θ + y sin θ = p, we obtain
cos θ = – \(\frac{\sqrt{3}}{2}\), sin θ = – \(\frac{1}{2}\) and p = 1
Since the values of sin θ and cos θ are negative, θ = π + \(\frac{\pi}{6}\) = \(\frac{7 \pi}{6}\).
Thus, the respective values of θ and p are \(\frac{7 \pi}{6}\) and 1.

Question 3.
Find the equation of the lines, which cut-off intercepts on the axes whose sum and product are 1 and -6, respectively.
Answer.
Let the intercepts form of line be \(\frac{x}{a}+\frac{y}{b}\) = 1,
then a + b = 1 and ab = – 6.
b = 1 – a and a (1 – a) = – 6
⇒ a² – a – 6 = 0
⇒ (a – 3) (a + 2) = 0
∴ a = 3, – 2

Case I:
If a = 3, then b = – \(\frac{6}{a}\)
= – \(\frac{6}{3}\) = – 2
∴ Equation of line is \(\frac{x}{3}+\frac{y}{-2}\) = 1
⇒ 2x – 3y – 6 = 0.

Case II:
If a = – 2, then b = \(\frac{-6}{-2}\) = 3
∴ Equation of line is \(\frac{x}{-2}+\frac{y}{3}\) = 1
⇒ 3x – 2y + 6 = 0.

Question 4.
What are the points on the y-axis whose distance from the line \(\frac{x}{3}+\frac{y}{4}\) = 1 is 4 units.
Answer.
Let (0, b) be the point on they-axis whose distance from line \(\frac{x}{3}+\frac{y}{4}\) = 1 is 4 units.
The given line can be written as 4x + 3y – 12 = 0 ……………..(i)
On comparing equation (i) to the general equation of line Ax + By + C = 0,we obtain A = 4, B = 3 and C = – 12.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by
d = \(\)
Therefore, if (0, b) is the point on the y-axis whose distance from line \(\frac{x}{3}+\frac{y}{4}\) = 1 is 4 units, then:
4 = \(\frac{|4(0)+3(b)-12|}{\sqrt{4^{2}+3^{2}}}\)

4 = \(\frac{|3 b-12|}{5}\)
⇒ 20 = |3b – 12|
20 = ±(3b – 12)
20 = (3b – 12) or 20 = – (3b – 12)
3b = 20 + 12 or 3b = – 20 + 12
b = \(\frac{32}{3}\) or b = \(\frac{8}{3}\).
Thus, the required points are (o, \(\frac{32}{3}\)) and (o, \(\frac{8}{3}\)).

Question 5.
Find perpendicular distance from the origin of the line joining the points (cos θ, sin θ) and (cos Φ, sin Φ).
Answer.
Equation of the line joining the points (cos θ, sin θ) are (cos Φ, sin Φ) is given by

Question 6.
Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.
Answer.
The point of intersection (x₁, y₁) of the lines x – 7y + 5 = 0 and 3x + y = 0 is obtained by solving these equations.
Putting y = – 3x in x – 7y + 5 = 0
⇒ x – 7 (- 3x) + 5 = 0,
⇒ x + 21x + 5 = 0
⇒ 22x + 5 = 0
⇒ x = \(\frac{5}{22}\)
Also, y = \(\frac{15}{12}\)
⇒ (x₁, y₁) = \(\left(\frac{-5}{22}, \frac{15}{12}\right)\)
Let a line parallel to y-axis through the point (x₁, y₁) is x = x₁
Here x₁ = – \(\frac{5}{22}\)
∴ The equation of the line parallel to y-axis passing through the point of intersection (x₁, y₁) of given lines is x = – \(\frac{5}{22}\) or 22x + 5 = 0.

Question 7.
Find the equation of a line drawn perpendicular to the line \(\frac{x}{4}+\frac{y}{6}\) = 1 through the point, where It meets the y-axis.
Answer.
Given equation of line is \(\frac{x}{4}+\frac{y}{6}\) = 1

\(\frac{3 x+12}{12}\) = 1

3x + 2y = 12 ……………..(i)
If line (i) meet the Y-axis, then put x = 0 in eq. (i), we get
0 + 2y = 12
⇒ y = 6
∴ Point is (0, 6).
Slope of line (i) is, m₁ = – \(\frac{3}{2}\)
Slope of line perpendicular to line (i) is,
m₂ = – \(-\frac{1}{m_{1}}=\frac{-1}{(-3 / 2)}=\frac{2}{3}\)

Now, equation of line having slope and passing through (0, 6) is given by
y – y₁ = m (x – x₁)
⇒ y – 6 = \(\frac{2}{3}\) (x – 0)
⇒ 3y – 18 = 2x
⇒ 2x – 3y + 18 = 0
Which is required equation of line.

Question 8.
Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k= 0.
Answer.

Let ABC be the triangle, whose sides are
AB: x – k = 0 ……………..(i)
BC: y – x = 0 ………………(ii)
and AC: x + y = 0 ………………(iii)
On solving eqs. (i) and (iii), we get Coordinates of A i.e.,(k, – k)
On solving eqs. (i) and (ii), we get Coordinates of B i.e.,(k, k)
On solving eqs. (ii) and (iii), we get Coordinates of C i.e. (0, 0)
∴ Area of ∆ABC = \(\frac{1}{2}\) {x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)
= \(\frac{1}{2}\) {k (k – 0) + k (0 + k) + 0 (- k – k)}
[∵ (x₁, y₁) = (k,- k), (x₂, y₂) = (k, k) and (x₃, y₃) = (0, 0)]
= \(\frac{1}{2}\) {k² + k²} = \(\frac{1}{2}\) k²
= k²

Question 9.
Find the value of p so that the three lines 3x + y – 2 = 0 px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.
Answer.
The equations of the given lines are
3x + y – 2 = 0 …………(i)
px + 2y – 3 = 0 …………..(ii)
2x – y – 3 = 0 ……………..(iii)
On solving equations (i) and (iii), we obtain x = 1 and y = – 1.
Since these three lines may intersect at one point, the point of intersection of lines (i) and (iii) will also satisfy line (ii). p(1) + 2(- 1) – 3 = 0
p – 2 – 3 = 0
⇒ p = 5
Thus, the required value of p is 5.

Question 10.
If three lines whose equations are y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ are concurrent, then show that m₁ (c₂ – c₃) + m₂ (c₃ – c₁) + m₃ (c₁ – c₂) = 0.
Answer.
The equations of the given lines are
y = m₁x + c₁ ………………(i)
y = m₂x + c₂ ……………..(ii)
y = m₃x + c₃ ………………(iii)
On subtracting equation (i) from (ii), we obtain 0 = (m₂ – m₁) x + (c₂ – c₁)
⇒ (m₁ – m₂)x = c₂ – c₁
⇒ x = \(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}\)

On substituting this value of x in eq. (i), we obtain

y = \(m_{1}\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}\right)+c_{1}=\frac{m_{1} c_{2}-m_{1} c_{1}}{m_{1}-m_{2}}+c_{1}\)

= \(\frac{m_{1} c_{2}-m_{1} c_{1}+m_{1} c_{1}-m_{2} c_{1}}{m_{1}-m_{2}}=\frac{m_{1} c_{2}-m_{2} c_{1}}{m_{1}-m_{2}}\)

∴ \(\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}, \frac{m_{1} c_{2}-m_{2} c_{1}}{m_{1}-m_{2}}\right)\) is the point of intersection of lines (i) and (ii).

It is given that lines (i), (ii) and (iii) are concurrent.
Hence the point of intersection of lines (i) and (ii) will also satisfy equation (iii).

\(\frac{m_{1} c_{2}-m_{2} c_{1}}{m_{1}-m_{2}}=m_{3}\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}\right)+c_{3}\) \(\frac{m_{1} c_{2}-m_{2} c_{1}}{m_{1}-m_{2}}=\frac{m_{3} c_{2}-m_{3} c_{1}+c_{3} m_{1}-c_{3} m_{2}}{m_{1}-m_{2}}\)

m₁c₂ – m₂c₁ – m₃c₁ – c₃m₁ + c₃m₂ = 0

m₁ (c₂ – c₃) + m₂ (c₃ – c₁) + m₃ (c₁ – c₂) = 0.
Hence proved.

Question 11.
Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x – 2y = 3.
Answer.
Let the slope of the required line be m₁.
The given line can be written as y = \(\frac{1}{2} x-\frac{3}{2}\) which is of the form y = mx + c.
Slope of the given line = m₂ = \(\frac{1}{2}\)
It is given that the angle between the required line and line x – 2y = 3 is 45°
We know that if θ is the acute angle between lines l₁ and l₂ with slopes m₁ and m₂ respectively, then

Case I : m₁ = 3
The equation of the line passing through (3, 2) and having a slope of 3 is:
y – 2 = 3 (x – 3)
y – 2 = 3x – 9
3x – y = 7

Case II: m₁ = – \(\frac{1}{3}\).
The equation of the line passing through (3, 2) and having a slope of – \(\frac{1}{3}\) is
y – 2 = – \(\frac{1}{3}\) (x – 3)
3y – 6 = – x + 3
x + 3y = 9.
Thus, the equations of the lines are 3x – y = 7 and x + 3y = 9.

Question 12.
Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Answer.

The given lines are
2x – 3y = – 1 …………….(i)
4x + 7y = 3 ……………(ii)
Multiplying eq. (i) by 2
4x – 6y = – 2 ………….(iii)
Subtracting eq. (iii) from eq. (ii), we get
13y = 5
⇒ y = \(\frac{13}{5}\)
Putting the value of y in eq. (i),
2x – \(\frac{3 \times 5}{13}\) = – 1
2x = – 1 + \(\frac{15}{13}=\frac{2}{13}\)
x = \(\frac{1}{13}\)
∴ Given lines intersect at P(\(\frac{1}{13}\), \(\frac{5}{13}\))
PA and PB are the lines that make equal intercepts on the axes.
They make angles of 135° with positive direction of x-axis.
Their slopes are tan 135° and tan 45° i.e., – 1 and 1 respectively.
∴ Equation of PA is y – \(\frac{5}{13}\) = (- 1) × (x – \(\frac{1}{13}\))
or 13y – 5 = – 13x + 1
13x + 13y – 6 = 0
Similarly, equation of PB is y – \(\frac{5}{13}\) = 1 × (x – \(\frac{5}{13}\))
⇒ 13y – 5 = 13x – 1
13x – 3y + 4 = 0.

Question 13.
Show that the equation of the line passing through the origin and nisildng an angle θ with the line y = mx + c is
\(\frac{y}{x}=\frac{m \pm \tan \theta}{1 \mp m \tan \theta}\)
Answer.
Slope of line y = mx + c is m.
Let M be the slope of required line, then
tan θ = \(\left|\frac{M-m}{1+m M}\right|=\pm\left(\frac{M-m}{1+M m}\right)\)

Case I:
Taking ‘+‘ sign,
tan θ = \(\frac{M-m}{1+m M}\)
Then, tan θ + Mm . tan θ = M – m

Question 14.
In what ratio, the line joining (- 1, 1) and (5, 7) is divided by the line x + y = 4?
Answer.
The equation of the line joining the points (- 1, 1) and (5, 7) is given by
y – 1 = \(\frac{7-1}{5+1}\) (x + 1)
= \(\frac{6}{6}\) (x + 1)
x – y + 2 = 0 ………….(i)
The equation of the given line is x + y – 4 = 0 ……………..(ii)
The point of intersection of lines (i) and (ii) is given by x = 1 and y = 3.
Let point (1, 3) divide the line segment joining (- 1, 1) and (5, 7) in the ratio 1 : k. Accordingly, by section formula.
(1, 3) = \(\left(\frac{k(-1)+1(5)}{1+k}, \frac{k(1)+1(7)}{1+k}\right)\)

(1, 3) = \(\left(\frac{-k+5}{1+k}, \frac{k+7}{1+k}\right)\)

\(\frac{-k+5}{1+k}\) = 1

∴ – k + 5 = 1 + k
⇒ 2k = 4
⇒ k = 2.
Thus, the line joining the points (- 1, 1) and (5, 7) is divided by line x + y = 4 in the ratio 1 : 2.

Question 15.
Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.
Answer.
The given lines are
2x – y = 0 ………………..(i)
4x + 7y + 5 = 0 ………………..(ii)
A(1, 2) is a point on line (i).
Let B be the point of intersection of lines (i) and (ii).
On solving equations (i) and (ii), we obtain x = – \(\frac{5}{18}\) and y = – \(\frac{5}{9}\).
∴ Coordnates of point B = (- \(\frac{5}{18}\), – \(\frac{5}{9}\))

Thus, the required distance is \(\frac{23 \sqrt{5}}{18}\) units.

Question 16.
Find the direction in which a straight line must be drawn through the point (- 1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.
Answer.
Any line passing through P(- 1, 2) is
y – 2 = m(x + 1)
where m is its slope
y = mx + m + 2 ……………(i)
Putting the value of ‘y’ in x + y = 4, we get
x + mx +m + 2 = 4
⇒ (1 + m) x = 4 – 2 – m
⇒ x = \(\frac{2-m}{1+m}\)
From eq.(i),
y = m (\(\frac{2-m}{1+m}\)) + m + 2
Now from the diagram, it is clear that the point of intersection of the other two legs of the right triangle having AB as the hypotenuse can be either P or Q.

Case I: When ∠APB is taken,
The perpendicular sides in ∠APB are AP and PB.
Now, side PB is parallel to x-axis and at a distance of 1 unit above x-axis.
So, equation of PB is y = 1 or y – 1 = 0.
The side AP is parallel to y-axis and at a distance of 1 unit on the right of y-axis.
So, equation of AP is x = 1 or x – 1 = 0.

Case II: When ∠AQB is taken.
The perpendicular sides in ∠AQB are AQ and QB.
Now, side AQ is parallel to x-axis and at a distance of 3 units above x – axis.
So, equation of AQ is y = 3 or y – 3 = 0.
The side QB is parallel to y-axis and at a distance of 4 units on the left of y-axis.
So, equation of QB is x = – 4 or x + 4 = 0.
Hence, the equation of the legs are: x = 1, y = 1 or x = – 4, y = 3.

Question 18.
Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Answer.
Let AB be the line x + 3y = 7 and the image P(3, 8) of P(3, 8) be Q(x₁, y₁)middle point at PQ

M \(\left(\frac{x_{1}+3}{2}, \frac{y_{1}+8}{2}\right)\) lies on AB.

∴ \(\left(\frac{x_{1}+3}{2}\right)+3\left(\frac{y_{1}+8}{2}\right)\) = 7
x₁ + 3 + 3y₁ + 24 = 14
⇒ x₁ + 3y₁ + 13 = 0 ……………….(i)
slope of AB = – \(\frac{1}{3}\),
Slope of PQ = \(\frac{y_{1}-8}{x_{1}-3}\)
AB ⊥ PQ

\(\left(-\frac{1}{3}\right)\left(\frac{y_{1}-8}{x_{1}-3}\right)\) = – 1

⇒ y₁ – 8 = 3 (x₁ – 3) = 3x₁ – 9
⇒ y₁ = 3x₁ – 1 ………….(ii)
Putting the value of y₁ in eq. (1), we get
x₁ + 3 (3x₁ – 1) + 13 = 0
10x₁ + 10 = 0
x₁ = – 1
Putting the value of x₁ in (ii), we get
y₁ = – 3 – 1 = – 4
∴ The point Q, the image of P is (- 1, 4).

Question 19.
If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.
Answer.
The equations of the given lines are
y = 3x + 1
2y = x + 3 …………..(ii)
y = mx + 4 …………….(iii)
Slope of line (i), m₁ = 3.
Slope of line (ii), m₂ = \(\frac{1}{2}\)
Slope of line (iii), m₃ = m.
It is given that lines (i) and (ii) are equally inclined to line (iii).
This means that the angle between lines (i) and (iii) equals the angles between lines (ii) and (iii).

Question 20.
If sum of the perpendicular distances of a variable point P(x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.
Answer.
The equations of the given lines are
x + y – 5 = 0 …………….(i)
3x – 2y + 7 = 0 ………………(ii)
The perpendicular distances of P(x, y) from lines (j) and (ii) are respectively given by
d₁ = \(\frac{|x+y-5|}{\sqrt{(1)^{2}+(1)^{2}}}\) and

d₂ = \(\frac{|3 x-2 y+7|}{\sqrt{(3)^{2}+(-2)^{2}}}\)

i.e., d₁ = \(\frac{|x+y-5|}{\sqrt{2}}\)

d₂ = \(\frac{|3 x-2 y+7|}{\sqrt{13}} .\)

It is given that d₁ + d₂ = 10

∴ \(\frac{|x+y-5|}{\sqrt{2}}+\frac{|3 x-2 y+7|}{\sqrt{13}}\) = 10

⇒ √13 |x + y – 5| + √2 |3x – 2y + 7| – 10√26 = 0
⇒ √13 (x + y – 5) + √2 (3x – 2y + 7) – 10√26 = 0
[Assuming (x + y – 5) and (3x – 2y + 7) are positive]
⇒ √13x + √13y – 5√13 + 3√2x – 2√2y + 7√2 – 10√26 = 0
⇒ x (√13 + 3√2) + y (√13 – 2√2) + (7√2 – 5√13 – 10√26) = 0
which is the equation of a line.
Similarly, we can obtain the equation of line for any signs of (x + y – 5) and (3x – 2y + 7).
Thus, point P must move on a line.

Question 21.
Find equation of the line which is equidistant from parallel lines 9x+6y – 7 = 0 and 3x + 2y + 6 = 0.
Answer.
The given parallel lines are
9x + 6y – 7 = 0 …………….(i)
and 3x + 2y + 6 = 0 …………….(ii)
Multiplying (ii) by 3, we get
9x + 6y + 18 = 0 ………………….(iii)
Let the equations of (ii) line parallel to the lines (i) and (iii) is
9x + 6y + c = 0 …………..(iv)
Distance between (i) and (iv)
= \(\frac{|-7-c|}{\sqrt{9^{2}+6^{2}}}=\frac{|7+c|}{\sqrt{117}}\)
Distance between (iii) and (iv)
= \(\frac{|18-c|}{\sqrt{9^{2}+6^{2}}}\)
The third line being equidistant from the given two lines.
\(\frac{|7+c|}{\sqrt{117}}=\frac{|c+18|}{\sqrt{117}}\) or 2c = 11 or c = \(\frac{11}{2}\)
Putting this values of c in eq. (iv), we get
9x + 6y + \(\frac{11}{2}\) = 0
or 18x + 12y + 11 = 0
which is the equation of required line.

Question 22.
A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.
Answer.
In the figure, PA is the incident ray and AR is the reflected ray, which makes an angle 0 from the X-axis.

It is clear from the figure that AS ⊥ OX
It means AS bisect the ∠PAR.
Then, ∠PAS = ∠RAS
⇒ ∠RAX = ∠PAO = θ (let)
⇒ ∠XAP = 180° – θ
Slope of AR = tan θ
= \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

= \(\frac{3-0}{5-k}\) …………….(i)
[where, point A is (k, 0)]
Slope of AP = tan (180 – θ)
= – tan θ
= \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2-0}{1-k}\) ……………(ii)
From eqs. (i) and (ii), we get
\(\frac{3}{5-k}=-\frac{2}{1-k}\)

⇒ 3 – 3k = – 10 + 2k
⇒ 5k = 13
⇒ k = \(\frac{13}{5}\)
Hence, the coordinates of A are (\(\frac{13}{5}\). 0).

Question 23.
Prove that the product of the lengths of the perpendiculars drawn from the points (\(\sqrt{a^{2}-b^{2}}\), 0) and (- \(\sqrt{a^{2}-b^{2}}\), 0) to the line \(\frac{x}{a}\) cos θ + \(\frac{y}{4}\) sin θ = 1 is b².
Answer.
The equation of the given line is \(\frac{x}{a}\) cos θ + \(\frac{y}{4}\) sin θ = 1
or bx cos θ + ay sin θ – ab = 0 ……………….(i)
Length of the perpendicular from point (\(\sqrt{a^{2}-b^{2}}\), 0) to the line (i) is
P₁ = \(\frac{\mid b \cos \theta\left(\sqrt{\left.a^{2}-b^{2}\right)}+a \sin \theta(0)-a b \mid\right.}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)

= \(\frac{\left|b \cos \theta \sqrt{a^{2}-b^{2}}-a b\right|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\) ……………….(ii)
Length of the perpendicular from point (- \(\sqrt{a^{2}-b^{2}}\), 0) to line (ii) is
P₁ = \(\frac{b \cos \theta\left(-\sqrt{a^{2}-b^{2}}\right)+a \sin \theta(0)-a b}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)

= \(\frac{\left|b \cos \theta \sqrt{a^{2}-b^{2}}+a b\right|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)

On multiplying equations (ii) and (iii), we obtain

Question 24.
A person standfing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = O and
3x + 4y – 5 = 0 wants to reach the path whose equation Is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.
Answer.
The equations of the given lines are
2x – 3y + 4 = 0
3x + 4y – 5 = 0 …………..(ii)
6x – 7y + 8 = 0 …………(iii)
The person is standing at the junction of the paths represented by lines (i) and (ii).
On solving equations (i) and (ii), we obtain x = – \(\frac{1}{17}\) and y = \(\frac{22}{17}\)
Thus, the person is standing at point (- \(\frac{1}{17}\), \(\frac{22}{17}\))
The person can reach path (iii) in the least time if he walks along the perpendicular line to (iii) from point (- \(\frac{1}{17}\), \(\frac{22}{17}\)).
Slope of the line (iii) = \(\frac{6}{7}\).
∴ Slope of the line perpendicular to line (iii) = \(-\frac{1}{\left(\frac{6}{7}\right)}=-\frac{7}{6}\)
The equation of the line passing through and having a slope (- \(\frac{1}{17}\), \(\frac{22}{17}\)) and having a slope of – \(\frac{7}{6}\) is given by
\(\left(y-\frac{22}{17}\right)=-\frac{7}{6}\left(x+\frac{1}{17}\right)\)
6 (17y – 22) = – 7 (17x + 1)
102y – 132 = – 119x – 7
119x + 102y = 125
Hence, the path that the person should follow is 119x + 102y = 125.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 10 Straight Lines Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3

Question 1.
Reduce the following equations into slope-intercept form and find their slopes and the y-intercepts.
(i) x + 7y = 0
(ii) 6x + 3y – 5 = 0
(iii) y = 0
Answer.
(i) The given equation is x + 7y = 0
It can be written as
y = – \(\frac{1}{7}\) x + 0 …………….(i)
This equation is of the form y = mx + c, where m = – \(\frac{1}{7}\) and c = 0.
Therefore, equation (i) is in the slope-intercept form, where the slope and the y-intercept are – \(\frac{1}{7}\) and 0 respectively.

(ii) The given equation is 6x + 3y – 5 = 0.
It can be wrirtenas y = \(\frac{1}{3}\) (- 6x + 5)
y = – 2x + \(\frac{5}{3}\)
This equation is of the form y = mx + c, where m = – 2 and c = \(\frac{5}{3}\)
Therefore, equation (ii) is in the slope-intercept form, where the slope and the y-intercept are – 2 and \(\frac{1}{3}\) respectively.

(iii) The given equation is y = 0.
It can be written as y = 0.x + 0 …………..(iii)
This equation is of the form y = mx + c, where m = 0 and c = 0.
Therefore, equation (iii) is in the slope-intercept form, where the slope and the y-intercept are 0 and 0 respectively.

Question 2.
Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0
(ii) 4x – 3y = 6
(iii) 3y + 2 = 0
Answer.
(i) The given equation is 3x + 2y -12 = 0
It can be written as 3x + 2y = 12
\(\frac{3 x}{12}+\frac{2 y}{12}\) = 1

i.e., \(\frac{x}{4}+\frac{y}{6}\) = 1

This equation is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1, where a = 4 and b = 6.
Therefore, equation (i) is in the intercept form, where the intercepts on the x and y axes are 4 and 6 respectively.

(ii) The given equation is 4x – 3y = 6
It can be written as \(\frac{4 x}{6}-\frac{3 y}{6}\) = 1
\(\frac{2 x}{3}-\frac{y}{2}\) = 1

\(\frac{x}{\left(\frac{3}{2}\right)}+\frac{y}{(-2)}\) = 1
This equation is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1, where a = \(\frac{3}{2}\) and b = – 2.
Therefore, equation (ii) is in the intercept form, where the intercepts on the x and y – axes are \(\frac{3}{2}\) and – 2 respectively.

(iii) The given equation is 3y +2 = 0.
It can be written as 3y = – 2
i.e., \(\frac{y}{\left(-\frac{2}{3}\right)}\) = 1 ……………….(iii)
This equation is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1,where a = 0 and b = – \(\frac{2}{3}\).
Therefore, equation (iii) is in the intercept form, where the intercept on the y-axis is – \(\frac{2}{3}\) and it has no intercept on the x-axis.

Question 3.
Reduce the following equations into the nornml form. Find their perpendicular distance from the origin and nngle between perpendicular and positive direction of x-axis.
(i) x – √3y + 8 = 0
(ii) y – 2 = 0
(iii) x – y = 4
Answer.
(i) x – √3y + 8 = 0 or
x – √3y = – 8 or
– x + √3y = 8
Put r cos ω = – 1, r sin ω = √3
Squaring and adding r² = 1 + 3 = 4
∴ r = 2
cos ω = – \(\frac{1}{2}\), sin ω = \(\frac{\sqrt{3}}{2}\) is
∴ ω = 120 = \(\frac{2 \pi}{3}\)
∴ x cos w + y sin w = \(\frac{8}{2}\) = 4
∴ p = 4 and ω = \(\frac{2 \pi}{3}\)

(ii) y – 2 = 0
Comparing with Ax + By = C
A = 0, B = 1, r = \(\sqrt{0+1^{2}}\) = 1
cos ω = 0, sin ω = 1
∴ y = 2
⇒ x cos ω + y sin ω = 2
∴ ω = \(\frac{\pi}{2}\), p = 2

(iii) x – y = 4
Put r cos ω = 1, r sin ω = – 1
r = \(\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}\)
cos ω = \(\frac{1}{\sqrt{2}}\), sin ω = – \(\frac{1}{\sqrt{2}}\)
ω = 360° – 45°= 315°
Dividing eq. (i) by √2
\(\frac{1}{\sqrt{2}}\) x + (- \(\frac{1}{\sqrt{2}}\) y) = \(\frac{4}{\sqrt{2}}\) = 2√2
Normal form of the given line
x cos 315 + y sin 315 = 2√2
ω = 315°, p = 2√2.

Question 4.
Find the distance of the point (- 1, 1) from the line 12 (x + 6) = 5 (y – 2).
Answer.
The given equation of the line is 12(x + 6) = 5 (y – 2).
⇒ 12x + 72 = 5y – 10.
⇒ 12x – 5y + 82 = 0 ………………(i)
On comparing equation (i) with general equation of line Ax + By + C = 0, we obtain A = 12, B = – 5, and C = 82.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by
d = \(\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}\)
The given point is (x₁, y₁) = (- 1, 1).
Therefore the distance of point (- 1, 1) from the given line
= \(\frac{|12(-1)+(-5)(1)+82|}{\sqrt{(12)^{2}+(-5)^{2}}}\) units

= \(\frac{|-12-5+82|}{\sqrt{169}}\) units

= \(\frac{|65|}{13}\) units = 5 units.

Question 5.
Find the points on the x-axis, whose distances from the line \(\) = 1 are 4 units.
Answer.
The given equation of line is \(\) = 1
or 4x + 3y -12 = 0 …………………(i)
On comparing equation (i) with general equation of line Ax + By + C = 0, we obtain A = 4, B = 3 and C = – 12.
Let (a, 0) be.the point on the x-axis whose distance from the given line is 4 units.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

d = \(\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}\)

Therefore, 4 = \(\frac{|4 a+3 \times 0-12|}{\sqrt{4^{2}+3^{2}}}\)

4 = \(\frac{|4 a-12|}{5}\)

⇒ |4a – 12| = 20
⇒ (4a – 12) = 20 or ± (4a – 12) = 20
⇒ (4a – 12) = 20 or – (4a – 12) = 20
⇒ 4a = 20 + 12 or 4a = – 20 + 12
⇒ a = 8or – 2
Thus, the required points on the x-axis are (- 2, 0) and (8, 0).

Question 6.
Find the distance between parallel lines
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
(ii) l(x + y) + p = 0 and l(x + y) – r = 0
Answer.
It is known that the distance (d) between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is given by d = \(\frac{\left|C_{1}-C_{2}\right|}{\sqrt{A^{2}+B^{2}}}\)

(i) The given parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.
Here, A = 15, B = 8, C₁ = – 34 and C₂ = 31.
Therefore, the distance between the piralle1 lines is
d = \(\frac{\left|C_{1}-C_{2}\right|}{\sqrt{A^{2}+B^{2}}}\)

= \(\frac{|-34-31|}{\sqrt{(15)^{2}+(8)^{2}}}\) units

= \(\frac{|-65|}{17}\) units = \(\frac{65}{17}\) units

(ii) The given parallel lines are l(x + y) + p =0 and l(x + y) – r = 0.
lx + ly + p = 0 and lx + ly – r = 0
Here, A = l, B = l, C₁ = p and C₂ = – r.
Therefore, the distance between the parallel lines is

d = \(\frac{\left|C_{1}-C_{2}\right|}{\sqrt{A^{2}+B^{2}}}\)

= \(\frac{|p+r|}{\sqrt{l^{2}+l^{2}}}\) units

= \(\frac{|p+r|}{\sqrt{2 l^{2}}}\) units

= \(\frac{|p+r|}{l \sqrt{2}}\) units

= \(\frac{1}{\sqrt{2}}\left|\frac{p+r}{l}\right|\) units.

Question 7.
Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (- 2, 3).
Answer.
The equation of the given line is 3x – 4y + 2 = 0
y = \(\)

or y = \(\), which is of the form y = mx + c
∴ Slope of the given line = \(\frac{3}{4}\)
It is known that parallel lines have the same slope.
∴ Slope of the other line = m = \(\frac{3}{4}\)
Now, the equation of the line that has a slope of \(\frac{3}{4}\) and passes through the point (- 2, 3) is
(y – 3) = \(\frac{3}{4}\) {x – (- 2)}
4y – 12 = 3x + 6
i.e., 3x – 4y + 18 = 0.

Question 8.
Find equation of the line perpendicular to the line x ly + 5=0 and having x intercept 3.
Answer.
Given equation of line is
x – 7y + 5 = 0 …………..(i)
The slope of a line is m₁ = \(\frac{1}{7}\)
∴ The slope of a perpendicular line is m = – 7.
Let the required perpendicular line be y = mx + c
y = – 7x + c [∵ m = – 7] ……………(i)
Since, this line intercept the X-axis at 3.
∴ Eq. (i) passes through point (3, 0).
∴ 0 = – 7(3) + c
⇒ c = 21
On putting c = 21 in eq. (i),we get
y = – 7x + 21
7x + y = 21

Question 9.
Find angles between the lines √3x + y = 1 and x + √3y =1.
Answer.
Given lines be √3x + y = 1 or y = – √3x + 1 …………(i)
and x + √3y = 1
or y = \(-\frac{1}{\sqrt{3}} x-\frac{1}{\sqrt{3}}\) ……………(ii)
∴ Slope of line (i) = – √3 = m₁ (say)
and Slope of line (ii) = – \(\frac{1}{\sqrt{3}}\) = m₂ (say)
If θ is the angle between be lines (i) and (ii) then

tan θ = \(\frac{\left|m_{1}-m_{2}\right|}{1+m_{1} m_{2}}=\frac{\left|-\sqrt{3}+\frac{1}{\sqrt{3}}\right|}{1+(-\sqrt{3}) \times\left(-\frac{1}{\sqrt{3}}\right)}\)

= \(\frac{\left|\frac{-3+1}{\sqrt{3}}\right|}{1+1}=\frac{2}{\sqrt{3}} \cdot \frac{1}{2}\)

= \(\frac{1}{\sqrt{3}}\) = tan \(\frac{\pi}{6}\)

θ = \(\frac{\pi}{6}\).

Question 10.
The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 6, at right angle. Find the value of h.
Answer.
The slope of the line passing through points (h, 3) and (4, 1) is
m₁ = \(\frac{1-3}{4-h}=\frac{-2}{4-h}\)
The slope of line 7x – 9y – 19 = 0 or y = \(\frac{7}{9}\) x – \(\frac{19}{9}\) is m₂ = \(\frac{7}{9}\).
It is given that the two lines are perpendicular.
∴ m₁ × m₂ = – 1
⇒ 14 = 36 – 9h
⇒ 9h = 36 – 14
⇒ h = \(\frac{22}{9}\)
Thus, the value of h is \(\frac{22}{9}\).

Question 11.
Prove that the line through the point (x₁, y₁) and parallel to the line Ax + By + C = 0 is A(x – x) + B (y – y₁) = 0.
Solution:
The slope of line Ax + By + C = 0 or y = \(\left(\frac{-A}{B}\right) x+\left(\frac{-C}{B}\right)\) is m = \(-\frac{A}{B}\)
It is known that parallel lines have the same slope.
∴ Slope of the other line = m = \(-\frac{A}{B}\)
The equation of the line passing through point (x₁, y₁) and having a slope m = \(-\frac{A}{B}\) is
y – y₁ = m (x – x₁) = \(-\frac{A}{B}\) (x – x₁)
B(y – y₁) = – A (x – x₁)
A(x – x₁) + B(y – y₁) = 0.
Hence, the line through point (x₁, y₁) and parallel to line Ax + By + C = 0 is
A(x – x₁) + B(y – y₁) = 0.

Question 12.
Two lines passing through the point (2, 3) intersects each other at an Rngle of 60°. If slope of one line is 2, find equation of the other line.
Answer.
Let m be the slope of the other line.
Slope of line = 2
Angle between them = 60°
tan 60° = ± \(\frac{m-2}{1+2 m}\)

\(\frac{m-2}{1+2 m}\) = ± √3

(+ve) sign, m – 2 = √3 (1 + 2m) = √3 + 2√3m
or (2√3 – 1 )m = – 2 – √3
m = – \(\frac{(2+\sqrt{3})}{2 \sqrt{3}-1}\)
Equation of the line passing through the point (2, 3) with slope m is
y – 3 = \(\frac{(2+\sqrt{3})}{2 \sqrt{3}-1}\) (x – 2)

⇒ (2√3 – 1) y – 3 (2√3 – 1) = – (2 + √3) x + 2 (2 + √3)
⇒ (2√3 – 1) y – 6√3 + 3 = – (2 + √3) x + 4 + 2√3
⇒ (2 + √3) x + (2√3 – 1) y – 6√3 + 3 – 4 – 2√3 = 0
⇒ (2 + √3) x + (2√3 – 1) y – 8√3 – 1 = 0

Taking negative, sign \(\frac{m-2}{1+2 m}\) = – √3
m – 2 = – √3 (1 + 2m) = – √3 – 2√3m
(2√3 + 1) m = 2 – √3
m = \(\frac{2-\sqrt{3}}{1+2 \sqrt{3}}\)

So, equation of the line passing through (2, 3) with slope \(\frac{2-\sqrt{3}}{1+2 \sqrt{3}}\) is

y – 3 = \(\frac{2-\sqrt{3}}{1+2 \sqrt{3}}\) (x – 2)
⇒ (1 + 2√3) y – 3 (1 + 2√3) = (2 – √3) x – 2 (2 – √3)
⇒ (1 + 2√3) y – 3 – 6√3 = (2 – √3) x – 4 + 2√3
⇒ (2 – √3)x – (1 + 2√3) y – 1 + 8 = 0
Hence, required lines are
(√3 + 2)x + (2√3 – 1)y – 8√3 – 1 = 0
and (2 – √3) x – (1 + 2√3) + 8√3 – 1 = 0.

Question 13.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (- 1, 2).
Answer.
The right bisector of a line segment bisects the line segment at 90°.
The end-points of the line segment are given as A(3, 4) and B (- 1, 2).
Accordingly, mid – point of AB = \(\left(\frac{3-1}{2}, \frac{4+2}{2}\right)\) = (1, 3).
Slope of AB = \(\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2}\)
∴ Slope of the line perpendicular to AB = \(-\frac{1}{\left(\frac{1}{2}\right)}\) = – 2.
The equation of the line passing through (1, 3) and having a slope of – 2 is
(y – 3) = – 2 (x – 1)
y – 3 = – 2x + 2
2x + y = 5
Thus, the required ecuation of the line is 2x + y = 5.

Question 14.
Find the coordinates of the foot of perpendicular from the point (-1,3) to the line 3a; – 4y -16 = 0.
Answer.
The equation of the given line is 3x – 4y – 16 = 0
The equation of a line perpendicular to the given line is 4x + 3y + k = 0, where is a constant.
If this point passes through the point (- 1, 3) then
– 4 + 9 + k = 0
k = – 5
∴ The equation of a line passing through the point (- 1, 3) and perpendicular to the given line is
4x – 3y – 5 = 0
∴ The required foot of the perpendicular is the point of intersection of the lines
3x – 4y – 16 = 0 …………….(i)
and 4x + 3y – 5 = 0 …………….(ii)
Solving eqs. (i) and (ii) y cross – multiplication, we have

Hence, co-ordinates of first perpendicular are (\(\frac{68}{25}\), – \(\frac{49}{25}\)).

Question 15.
The perpendicular from the origin to the line y = mx + c meets it at the point (- 1, 2). Find the values of m and c.
Answer.
The given equation of line is y = mx + c;
It is given that the perpendicular from the origin meets the given line at (- 1, 2).
Therefore, the line joining the points (0, 0) and (- 1, 2) is perpendicular to the given line.
∴ Slope of the line joining (0, 0) and (- 1, 2) = \(\frac{2}{-1}\) = – 2
The slope of the given line is in.
∴ m × – 2 = – 1 [The two lines are perpendicular]
⇒ m = \(\frac{1}{2}\)
Since point (- 1, 2) lies on the given line, it satisfies the equation y = mx + c.
∴ 2 = m (- 1) + c
2 = \(\frac{1}{2}\) (- 1) + c
⇒ c = 2 + \(\frac{1}{2}\) = \(\frac{5}{2}\).
Thus, the respective values of m and c are \(\frac{1}{2}\) and \(\frac{5}{2}\).

Question 16.
If p and q are the lengths Qf perpendiculars from the origin to the lines x cos θ – y sin θ = k cos 2θ and x sec θ + y cosec θ = k respectively, prove that p² + 4q² = k².
Answer.
p = the length of perpendicular from origin (0, 0) to the line x cos θ – y sin θ – k cos 2θ = 0
Then, p = \(\frac{|0 \cdot \cos \theta+0 \cdot(-\sin \theta)-k \cos 2 \theta|}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}\)

[∵ the perpendicular distance from (x₁, y₁) tot the line ax + y + c = 0 is \(\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\)].

∴ p = \(\frac{k \cos 2 \theta}{1}\)
⇒ p = k cos 2θ ………………….(i)

and q = the length of perpendicular from (0, 0) to the line x sec θ + y cosec θ – k = 0
Then, q = \(\frac{|0 \cdot \sec \theta+0 \cdot \ {cosec} \theta-k|}{\sqrt{\sec ^{2} \theta+\ {cosec}^{2} \theta}}\)

q = \(\frac{k}{\sqrt{\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}}}=\frac{k}{\sqrt{\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta \cdot \cos ^{2} \theta}}}\)

= \(\frac{k \cdot \sin \theta \cos \theta}{1} \times \frac{2}{2}\)
[∵ sin² θ + cos² θ = 1]

q = \(\frac{k}{2}\) sin 2θ
2q = k sin 2θ ……………(ii)

On squaring eqs. (i) and (ii) and then adding, we get
p² + 4q² = k² cos² 2θ + k² sin² 2θ
= k² (cos² 2θ + sin² 2θ)
p² + 4q² = k²
[∵ cos² θ + sin² θ = 1]
Hence proved.

Question 17.
In the triangle ABC with vertices A(2, 3), B(4, 1) and C(1, 2), fInd the equation and length of altitude from the vertex A.
Answer.

Let AD be the altitude of triangle ABC from vertex A.
Accordingly, AD ⊥ BC
The equation of the line passing through point (2, 3) and having a slope of 1 is
(y – 3) = 1(x – 2)
⇒ x – y + 1 = 0
⇒ y – x = 1
Therefore, equation of the altitude from vertex A is y – x = 1.
Length of AD = Length of the perpendicular from A(2, 3) to BC.
The equation of BC is (y + 1) = \(\frac{2+1}{1-4}\) (x – 4)
⇒ (y + 1) = – 1 (x – 4)
⇒ y + 1 = – x + 4
⇒ x + y – 3 = 0 ……………… (i)
The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by
d = \(\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}\)
On comparing equation (i) to the general equation of line Ax + By + C = 0,
we obtain A = 1, B = 1 and C = – 3.
∴ Length of AD = \(\frac{|1 \times 2+1 \times 3-3|}{\sqrt{1^{2}+1^{2}}}\) units
= \(\frac{|2|}{\sqrt{2}}\) units

= \(\frac{2}{\sqrt{2}}\) units

= √2 units.

Thus, the equation and the length of the altitude from vertex A are y – x = 1 and √2 units respectively.

Question 18.
If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that
\(\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\).
Answer.
It is known that the equatìon of a line whose intercepts on the axes are a and b is
\(\frac{x}{a}+\frac{y}{b}\) = 1
or bx + ay = ab or bx + ay – ab = 0 …………….(i)
The perpendicular distance (d) of a line Ax + By + C = O from a point(x₁, y₁) is given by
d = \(\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}\)
On comparing equation (i) to the general equation of line Ax + By + C = 0, we obtain A = b, B = a, and C = – ab.
Therefore, if p is the length of the perpendicular from point (x₁, y₁) = (0, 0) to line (i), we obtain

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 10 Straight Lines Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2

Question 1.
Write the equations for the x and y-axes.
Answer.
The y-coordinate of every point on the x-axis is 0.
Therefore, the equation of the x – axis is y = 0.
The x-coordinate of every point on the y – axis is 0.
Therefore, the equation of the y-axis is x = 0.

Question 2.
Find the equation of the line which passes through the point (- 4, 3) with slope \(\frac{1}{2}\).
Answer.
We know that the equation of the line passing through point (x₀, y₀), whose slope is m, is (y – y₀) = m (x – x₀).
Thus, the equation of the line passing through point (- 4, 3), whose slope is \(\frac{1}{2}\) is
(y – 3) = – (x + 4)
2 (y – 3) = x + 4
2y – 6 = x + 4
i.e., x – 2y + 10 = 0.

Question 3.
Find the equation of the line which passes through (0, 0) with slope m.
Answer.
We know that the equation of the line passing through point (x₀, y₀), whose slope is m, is
(y – y₀) = m (x – x₀)
Thus, the equation of the line passing through point (0, 0), whose slope is m is (y – 0) = m(x – 0)
i.e., y = mx.

Question 4.
Find the equation of the line which passes through (2, 2√3) and is inclined with the x-s’ at an angle of 75°.
Answer.
The slope of the line that inclines with the x-axis at an angle of 75° is m = tan75°
m = tan (45° + 30°)
= \(\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \tan 30^{\circ}}\)

= \(\frac{1+\frac{1}{\sqrt{3}}}{1-1 \cdot \frac{1}{\sqrt{3}}}\)

= \(\frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\)

We know that the equation of the line passing through point (x₀, y₀), whose slope is m, is (y – y₀) = m(x – x₀).
Thus, if a line passes though (2, 2√3) and inclines with the x-axis at an angle of 75°, then the equation of the line is given as
(y – 2√3) = (x – 2)
(y – 2√3) (√3 – 1) = (√3 + 1)(x – 2)
y(√3 – 1) – 2 √3 (√3 – 1) = x(√3 + 1) – 2 (√3 + 1)
(√3 + 1) x – (√3 – 1) y = 2√3 + 2 – 6 + 2√3
(√3 + 1) x – (√3 – 1) y = 4√3 – 4
i.e., (√3 + 1) x – (√3 – 1) y = 4(√3 – 1).

Question 5.
Find the equation of the line which intersects the x-axis at a distance of 3 units to the left of origin with slope – 2.
Answer.
It is known that if a line with slope m makes x-intercept d, then the equation of the line is given as y = m(x – d).
For the line intersecting the x-axis at a distance of 3 units to the left of the origin, d = – 3.
The slope of the line is given as m = – 2.
Thus, the required equation of the given line is
y = – 2 [x – (- 3)] = – 2x – 6
i.e., 2x + y + 6 = 0.

Question 6.
Find the equation of the line which intersects the y-axis at a distance of 2 units above the origin and makes an angle of 30° with the positive direction of the x-axis.
Answer.

The line intersect the 7-axis at distance of 2 units above the origin. It show that the line passes through (0, 2).
It makes an angle 30° with the positive direction of X – axis.
1 X
So, the slope of the line is = tan 30° = \(\frac{1}{\sqrt{3}}\)
Thus, the equation of straight line is
y – y₁ = m (x – x₁)
y – 2 = \(\frac{1}{\sqrt{3}}\) (x – 0)
\(\frac{x}{\sqrt{3}}\) – y + 2 = 0
x – √3y + 2√3 = 0.

Question 7.
Find the equation of the line which passes through the points (- 1, 1) and (2, – 4).
Answer.
Given points are A (x₁, y₁) = (- 1, 1) and B(x₂, y₂) = (2, – 4), then equation of line AB is
y – y₁ = \(\frac{y-{2}-y-{1}}{x-{2}-x-{1}}\) (x – x₁)
⇒ y – 1 = \(\frac{-4-1}{2+1}\) (x + 1)
[x₁ = – 1, y₁ = 1, x₂ = 2, y₂ = – 4]
⇒ y – 1 = – \(\frac{5}{3}\) (x + 1)
⇒ 3y – 3 = – 5x – 5
⇒ 5x + 3y + 2 = 0

Question 8.
Find the equation of the line which is at a perpendicular distance of 5 units from the origin and the angle made by the perpendicular with the positive x – axis is 30°.
Answer.
If p is the length of the normal from the origin to a line and A is the angle made by the normal with the positive direction of the x-axis, then the equation of the line is given by x cos A + y sin A = p
Here, p = 5 units and A = 30°
Thus, the required equation of the given line is x cos 30° + y sin 30° = 5
x \(\frac{\sqrt{3}}{2}\) + y . \(\frac{1}{2}\) = 5
i.e., √3x + y = 10.

Question 9.
The vertices of ∆PQR are P(2, 1), Q(- 2, 3) and R(4, 5). Find equation of the median through the vertex R.
Answer.
Since, median bisects the opposite sides i. e., S is the mid-point of PQ.
S = \(\left(\frac{\ddot{x}-{1}+x-{2}}{2}, \frac{y-{1}+y-{2}}{2}\right)\)

= \(\left(\frac{2-2}{2}, \frac{1+3}{2}\right)=\left(0, \frac{4}{2}\right)\)

= (0, 2)

y – y₁ = \(\frac{y-{2}-y-{1}}{x-{2}-x-{1}}\) (x – x₁)
⇒ y – 5 = \(\frac{2-5}{0-4}\) (x – 4)
[∵ x₁ = 2, y₁ = 1, x₁ = – 2, y₁ = 3]
⇒ y – 5 = \(\frac{-3}{-4}\) (x – 4)
4y – 20 = 3x – 12
⇒ 3x – 4y + 8 = 0

Question 10.
Find the equation of the line passing through (- 3, 5) and perpendicular to the line through the points (2, 5) and (- 3, 6).
Answer.
The slope of the line joining the points (2, 5) and (- 3, 6) is
m = \(\frac{6-5}{-3-2}=\frac{1}{-5}\)

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, slope of the line perpendicular to the line through the points (2, 5) and (- 3, 6)
= \(-\frac{1}{m}=-\frac{1}{\left(\frac{-1}{5}\right)}\) = 5

Now, the equation of the line passing through point (- 3, 5), whose slope is 5, is
(y – 5) = 5 (x + 3)
y – 5 = 5x + 15
i.e., 5x – y + 20 = 0.

Question 11.
A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1 : n. Find the equation of the line.
Answer.

Let the given points are A(1, 0) and B(2, 3).
Let the line PQ divide AB in the ratio 1 : n at R internally.
Then, coordinates of R = \(\left(\frac{1 \times x-{2}+n \times x-{1}}{1+n}, \frac{1 \times y-{2}+n \times y-{1}}{1+n}\right)\)

= \(\left(\frac{1 \times 2+n \times 1}{1+n}, \frac{1 \times 3+n \times 0}{1+n}\right)\)

[∵ x₁ = 1, y₁ = 0, x₂ = 2, y₂ = 3]

= \(\left(\frac{n+2}{n+1}, \frac{3}{1+n}\right)\)

Let the slope of the line is m.
Also, PQ ⊥ AB
∴ Slope of the line PQ × Slope of the line AB = – 1
⇒ m × \(\frac{y-{2}-y-{1}}{x-{2}-x-{1}}\) = – 1

⇒ m × \(\frac{3-0}{2-1}\) = – 1

[∵ x₁ = 1, y₁ = 0, x₂ = 2, y₂ = 2]

⇒ m × 3 = – 1
⇒ m = – \(\frac{1}{3}\)

Now, equation of line PQ y using y – y₀ = m (x – x₀)

y – \(\frac{3}{1+n}\) = \(\frac{-1}{3}\left(x-\frac{n+2}{n+1}\right)\)

[∵ R\(\left(\frac{n+2}{n+1}, \frac{3}{1+n}\right)\) = (x₁, y₁)

\(\frac{3(n+1) y-9}{1+n}=\frac{-x(n+1)+(n+2)}{n+1}\)

⇒ 3 (n + 1) y – 9 = – x (n + 1) + (n + 2)
⇒ x (n + 1) + 3 (n + 1) y = n + 2 + 9
⇒ x (n + 1) + 3 (n + 1) y = n + 11
Which is the required equation of line.

Question 12.
Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).
Answer.
The equation of a line in the intercept form is
\(\frac{x}{a}+\frac{y}{b}\) = 1 …………….(i)
Here, a and b are the intercepts on x and y axes respectively.
It is given that the line cuts off equal intercepts on both the axes.
This means that a = b.
Accordingly, equation (i) reduces to
\(\frac{x}{a}+\frac{y}{a}\) = 1
x + y = a ………………(ii)
Since the given line passes through point (2, 3), equation (ii) reduces to
2 + 3 = a
a = 5
On substituting the value of a in equation (ii), we obtain x + y = 5, which is the required equation of the line.

Question 13.
Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Answer.
The equation of a line in the intercept form is
\(\frac{x}{a}+\frac{y}{b}\) = 1 ……………(i)
Here, a and b are the intercepts on x and y axes respectively.
It is given that a + b = 9
⇒ b = 9 – a ………….. (ii)
From equations (i) and (ii), we obtain
\(\frac{x}{a}+\frac{y}{9-a}\) = 1 ………..(iii)

It is given that the line passes through point (2, 2).
Therefore, equation (iii) reduces to \(\frac{2}{a}+\frac{2}{9-a}\) = 1
⇒ \(2\left(\frac{1}{a}+\frac{1}{9-a}\right)\) = 1

⇒ \(2\left(\frac{9-a+a}{a(9-a)}\right)\) = 1

⇒ \(\frac{18}{9 a-a^{2}}\) = 1

⇒ 18 = 9a – a²
⇒ a² – 9a + 18 = 0
⇒ a² – 6a – 3a + 18 = 0
⇒ a (a – 6) – 3 (a – 6) = 0
⇒ (a – 6) (a – 3) = 0
⇒ a = 6 or a = 3
If a = 6 and b = 9 – 6 =3, then the equation of the line is \(\frac{x}{6}+\frac{y}{3}\) = 1
⇒ x + 2y – 6 = 0
If a = 3 and = 9 – 3 = 6, then the equation of the line is
\(\frac{x}{6}+\frac{y}{3}\) = 1
⇒ 2x + y – 6 = 0.

Question 14.
Find equation of the line through the point (0, 2) making an angle – with the positive x-axis. Also, find the equation of line 3 parallel to it and crossing the y-axis at a distance of 2 units below the origin.
Answer.

Slope of the line = tan \(\frac{2 \pi}{3}\)
tan (π – \(\frac{\pi}{3}\)) = tan \(\frac{\pi}{3}\) = – √3
Equation of the line AC with slope = – √3 and passing through the point D given by
y – 2 = – √3(x – 0)
⇒ √3x + y – 2 = 0
For beyond past, we take another line BD parallel to AC.
∴ Slope of the line BD = Slope of AC = – √3
BD is passing through the point D{0, – 2)
∴ Equation of BD is y + 2 = √3(x – 0)
⇒ √3x + y + 2 = 0
Hence, equation of AC and BD are,
√3x + y – 2 = 0 and √3x + y + 2 = 0.

Question 15.
The perpendicular from the origin to a line meets it at the point (- 2, 9), find the equation of the line.
Answer.
The slope of the line joining the origin (0, 0) and point (- 2, 9) is
m₁ = \(\frac{9-0}{-2-0}=-\frac{9}{2}\)

Accordingly, the slope of the line, perpendicular to the line joining the origin and point (- 2, 9) is
m₂ = \(-\frac{1}{m-{1}}=-\frac{1}{\left(-\frac{9}{2}\right)}=\frac{2}{9}\)

Now, the equation of the line passing through point (- 2, 9) and having a slope m2 is
(y – 9) = \(\frac{2}{9}\) (x + 2)
9y – 81 = 2x + 4
2x – 9y + 85 = 0.

Question 16.
The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.
Answer.
Since, L is linear function of C.
L = a + bC …………….(i)
For L = 124.942, C = 20
124.942 = a + 20b ………………(ii)
For L = 125.134 C = 110
∴ 125.134 = a + 1106 ……………(iii)
Subtracting eq. (ii) from eq. (iii), we get
0. 192 = 90b
b = \(\frac{0.192}{90}\) = 0.00213
From eq. (ii),
124.942 = a + 20 × 0.00213 = a + 0.0426
a = 124.942 – 0.0426 = 124.8994
Hence from eq. (i) L in terms of C is
L = 124.8994 + 0.00213C.

Question 17.
The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs. 14/litre and 1220 litres of milk each week at Rs. 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs. 17/litre?
Answer.
Let price and litre be denoted in ordered pair (x, y), where x denotes the ₹ per litre and y denotes the quantity of milk in litre.
Given, (14980) and (161220) are two points.
Let linear relations i. e., linear equation be
y – y₁ = \(\frac{y-{2}-y-{1}}{x-{2}-x-{1}}\) (x – x₁)

y – 980 = \(\frac{1220-980}{16-14}\) (x – 14)

y – 980 = \(\frac{240}{2}\) (x – 14)
[∵ x₁ = 14, y₁ = 980, x₂ = 16, y₂ = 1220]
⇒ y – 980 = 120 (x – 14)
⇒ y – 980 = 120x – 120 × 14
⇒ 120x – y = 1680 – 980
⇒ 120x – y = 700
When price JC = 17,
120 x 17 – y = 700
y = 2040 – 700 = 1340
He will sell weekly 1340 L milk at the rate of ₹ 172.

Question 18.
P(a, b) is the mid-point of a line segment between axes. Show that equation of the line is \(\frac{x}{a}+\frac{y}{b}\) = 2.
Answer.
Let AB be the line segment between the axes and let P(a, b) be its mid-point.

Let the coordinates of A and B be (0, y) and (x, 0) respectively.
Since P(a, b) is the mid-point of AB.
\(\left(\frac{0+x}{2}, \frac{y+0}{2}\right)\) = (a, b)

\(\left(\frac{x}{2}, \frac{y}{2}\right)\) = (a, b) and

\(\frac{x}{2}\) = a and \(\frac{y}{2}\) = b
x = 2a and y = 2b
Thus, the respectively coordinares of A and B are (0, 2b) and (2a, 0).
The equation of the line passing through points (0, 2b) and (2a, 0) is

(y – 2b) = \(=\frac{(0-2 b)}{(2 a-0)}\) (x – 0)

y – 2b = \(\frac{-2 b}{2 a}\) (x)

a (y – 2b) = – bx
ay – 2ab = – bx
i.e., bx + ay = 2ab
On dividing both sides by ab, we obtain
\(\frac{b x}{a b}+\frac{a y}{a b}=\frac{2 a b}{a b}\)

⇒ \(\frac{x}{a}+\frac{y}{b}\) = 2
Thus, the equation of the line is \(\frac{x}{a}+\frac{y}{b}\) = 2.

Question 19.
Point R (h, k) divides a line segment between the axes in the ratio 1 : 2. Find equation of the line.
Answer.

Let AB be the line segment between the axes such that point R(h, k) divides AB in the ratio 1 : 2.
Let the respective coordinates of A and B be (x, 0) and (0, y).
Since point R(h, k) divides AB in the ratio 1 : 2, according to the section formula,
(h, k) = \(=\left(\frac{1 \times 0+2 \times x}{1+2}, \frac{1 \times y+2 \times 0}{1+2}\right)\)

(h, k) = \(\left(\frac{2 x}{3}, \frac{y}{3}\right)\)

⇒ h = \(\frac{2 x}{3}\) and k = \(\frac{y}{3}\)
⇒ x = \(\frac{3 h}{2}\) and y = 3k

Therefore, the respective coordinates of A and B are (\(\frac{3 h}{2}\), o) and (0, 3k).

Now, the equation of line AB passing through points (\(\frac{3 h}{2}\), 0) and (0, 3k) is

(y – 0) = \(\frac{3 k-0}{0-\frac{3 h}{2}}\) (x – \(\frac{3 h}{2}\))

y = – \(\frac{2 k}{h}\) (x – \(\frac{3 h}{2}\))
hy = – 2kx + 3hk
i.e., 2kx + hy = 3hk
Thus, the required equation of the line is 2kx + hy = 3hk.

Question 20.
By using the concept of equation of a line, prove that the three points (3, 0), (- 2, – 2) and (8, 2) are collinear.
Answer.
In order to show that points (3, 0) (- 2, – 2) and (8, 2) are coimear, it suffices to show that the line passing through points (3, 0) and (- 2, – 2) also passes through point (8, 2).
The equation of the line passing through points (3, 0) and (- 2, – 2) is
(y – 0) = \(\frac{(-2-0)}{(-2-3)}\) (x – 3)

y = \(\frac{-2}{-5}\) (x – 3)
5y = 2x – 6
i.e., 2x – 5y = 6
It is observed that at x = 8 and y = 2,
5y = 2x – 6
i.e., 2x – 5y = 6
It is observed that at x = 8 and y = 2,
L.H.S. = 2 × 8 – 5 × 2
= 16 – 10 = 6 = R.H.S.
Therefore, the line passing through points (3, 0) and (- 2, – 2) also passes through point (8, 2).
Hence, points (3, 0), (- 2, – 2) and (8, 2) are collinear.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 10 Straight Lines Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.1

Question 1.
Draw a quadrilateral in the Cartesian plane, whose vertices are (- 4, 5), (0, 7), (5, – 5) and (- 4, – 2). Also find its area.
Answer.
Let ABCD be the given quadrilateral with vertices A (- 4, 5), B (0, 7), C (5 – 5), and D (- 4. – 2).
Then, by plotting A, B, C and D on the Cartesian plane and joining AB, BC, CD and DA the given quadrilateral can be drawn as

To find the area of quadrilateral ABCD, we draw one diagonal, say AC.
Accordingly, area (ABCD) = area (∆ABC) + area (∆ACD)
We know that the area of a triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) is \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
Therefore, area of ∆ABC
= \(\frac{1}{2}\) |- 4 (7 + 5) + 0(- 5 – 5) + 5 (5 – 7)| unit²
= \(\frac{1}{2}\) |- 4 (12) + 5 (- 2)| unit²
= \(\frac{1}{2}\) |- 48 – 10| unit²
= \(\frac{1}{2}\) |- 58| unit²
= \(\frac{1}{2}\) × 58 unit²
= 29 unit²
Area of MCD = \(\frac{1}{2}\) |- 4 (- 5 + 2) + 5(- 2 – 5) ± (- 4) (5 + 5)| unit²
= \(\frac{1}{2}\) |- 4 (- 3) + 5 (- 7) – 4(10)| unit²
= \(\frac{1}{2}\) |12 – 35 – 40| unit²
= \(\frac{1}{2}\) |- 63| unit²
= \(\frac{63}{2}\) unit²
Thus, area (ABCD) = (29 + \(\frac{63}{2}\)) unit²
= \(\frac{58+63}{2}\) unit²
= \(\frac{121}{2}\) unit².

Question 2.
The base of an equilateral triangle with side 2a lies along the y-axis such that the mid point of the base is at the origin. Find vertices of the triangle.
Answer.

Let BC be the base of a triangle which lies on Y
Y-axis and third vertex may be A(h, 0) or A’.
Since, ∆ABC is an equilateral, then AB = BC.
∴ AB² = BC²
⇒ (h – 0)² + (0 – a²) = (2a)²
[∵ distance between two points (x₁, y₁) and (x₂, y₂)
= \(\left.\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]\)
For distance AB, (x₁, y₁) = (a, 0), (x₂, y₂) = (0, a)
h² + a² = 4a²
h² = 3a²
h = ± √3 a [taking square root]
Hence, the vertices of triangle are (√3a, 0), (0, a), (0,- a) or (- √3, a) (0, a), (0, – a).

Question 3.
Find the distance between P(x₁, y₁) and Q(x₂, y₂) when:
(i) PQ is parallel to the y – axis
(ii) PQ is parallel to the x-aLg.
Answer.

(i) When PQ is parallel to the Y-axis, it means the x-coordinates of P and Q are same i.e.,
x₁ = x₂.
∴ Distance between two points
PQ = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{\left(x_{1}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{\left(y_{2}-y_{1}\right)^{2}}=\left|y_{2}-y_{1}\right|\)

(ii)

When PQ is parallel to X-axis, it means y-coordinates of P and Q are same i.e., y₂ = y₁
∴ Distance between two points PQ = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

= \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{1}-y_{1}\right)^{2}}\) [∵ y₂ = y₁]

= \(\sqrt{\left(x_{2}-x_{1}\right)^{2}}\)

= |x₂ – x₁|

Question 4.
Find a point on the x-axis, which is equidistance from the points (7, 6) and (3, 4).
Answer.
Let any point P on the X-axis is (x, 0) as for a point on X-axis y-coordinate is zero and the given points are A (7, 6) and B (3, 4).
Given, PA = PB
⇒ PA² = PB²
⇒ (x₁ – x₂)² + (y₁ – y₂)² = (x₁ – x₃)² + (y₁ – y₃)²
where x₁ = x, x₂ = 7, y₁ = 0, y₂ = 6, x₃ = 3, y₃ = 4
(x – 7)² + (0 – 6)² = (x – 3)² + (0 – 4)²
[∵ distance between two points = \(\left.\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\right]\)]
x² + 49 – 14x + 36 = x² + 9 – 6x + 16
⇒ – 14x + 6x = 25 – 36 – 49
⇒ – 8x = 25 – 85
⇒ Point P on X-axis = (\(\frac{15}{2}\), 0).

Question 5.
Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0, – 4) and B(8, 0).
Answer.

Given points are p (0, – 4) and Q (8, 0).
∴ x₁ = 0, y₁ = – 4, x₂ = 8, y₂ = 0
These points plotted in XY-plane are given below.
Mid-point of PQ is R = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

= \(\left(\frac{0+8}{2}, \frac{-4+0}{2}\right)\) = (4, – 2)

∴ Slope of the OR = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

= \(\frac{-2-0}{4-0}=\frac{-2}{4}=-\frac{1}{2}\)

[∵ x = 0, x₂ = 4, y₁ = 0, y₂ = – 2]

Question 6.
Without using the Pythagoras theorem show that the points (4, 4), (3, 5) and (- 1, – 1) are the vertices of a right angled triangle.
Answer.
In ∆ABC, we have
m₁ = Slope of AB = \(\frac{4-5}{4-3}\) = – 1 and
m₂ = Slope of AC = \(\frac{4-(-1)}{4-(-1)}\) = 1
Clearly, m₁ m₂ = – 1
This shows that AB is perpendicular to AC i. e.,
∠CAB = π/2
Hence, the given points are the vertices of a right-angled triangle.

Question 7.
Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
Answer.
The line OP makes an angle of 30° with y-axis measured anticlockwise.

So OP makes an angle of 90° + 30° = 120° with positive direction of x-axis.
So, slope of OP = tan 120° = tan (180° – 60°) = – tan 60° = – √3.

Question 8.
Find the value of x for which the points (x, – 1), (2, 1) and (4, 5) are collinear.
Answer.
Given, points A (x, – 1), B (2, 1) and C (4, 5) are collinear.
Here, x₁ = x, y₁ = – 1, x₂ = 2, y₂ = 1, x₃ = 4 and y₃ = 5
∴ Slope of AB = Slope of BC
⇒ \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y_{3}-y_{2}}{x_{3}-x_{2}}\)

⇒ \(\frac{1+1}{2-x}=\frac{5-1}{4-2}\)

⇒ \(\frac{2}{2-x}=\frac{4}{2}\)

⇒ 2 – x = 1
⇒ x = 2 – 1 = 1

Question 9.
Without using distance formula, show that points (- 2, – 1), (4, 0), (3, 3) and (- 3, 2) are vertices of a parallelogram.
Answer.
Let ABCD be a parallelogram, where vertices are

(x₁, y₁) → A (- 2, – 1),
(x₂, y₂) → B (4, 0),
(x₃, y₃) → C (3, 3) and
(x₄, y₄) → D (- 3, 2)
Mid-point of AC = \(\left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}\right)=\left(\frac{-2+3}{2}, \frac{-1+3}{2}\right)\)

= \(\left(\frac{1}{2}, \frac{2}{2}\right)=\left(\frac{1}{2}, 1\right)\)

[∵ mid-point of two points = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)]

Mid point of BD = \(\left(\frac{x_{2}+x_{4}}{2}, \frac{y_{2}+y_{4}}{2}\right)\)

= \(\left(\frac{4-3}{2}, \frac{0+2}{2}\right)=\left(\frac{1}{2}, 1\right)\)

∵ Mid-point of AC = Mid-point of BD

∴ ABCD is parallelogram.
Hence proved.

Question 10.
Find the angle between the xr-axis and the line joining the points (3, – 1) and (4, – 2).
Answer.
The slope of the line joining the points (3, – 1) and (4, – 2) is
m = \(\frac{-2-(-1)}{4-3}\)
= – 2 + 1 = – 1
Now, the inclination (θ) of the line joining the points (3, – 1) and (4, – 2) is given by tan θ = – 1
⇒ θ = (90° + 45°) = 135°
Thus, the angle between the x-axis and the line joining the points (3, – 1) and (4, – 2) is 135°.

Question 11.
The slope of a line is double of the slope of another line. If tangent of the angle between them is \(\frac{1}{3}\), find the slopes of the lines.
Answer.
Let m₁ and m be the slopes of the two given lines such that m₁ = 2m
We know that if 0 is the angle between the lines l₁ and l₂ with slopes m₁ and m₂ then
tan θ = \(\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|\)
It is given that the tangent of the angle between the two lines is \(\frac{1}{3}\).
∴ \(\frac{1}{3}=\left|\frac{m-2 m}{1+(2 m) \cdot m}\right|\)

\(\frac{1}{3}=\left|\frac{-m}{1+2 m^{2}}\right|\)

\(\frac{1}{3}=\frac{-m}{1+2 m^{2}}\) or

\(\frac{1}{3}=-\left(\frac{-m}{1+2 m^{2}}\right)=\frac{m}{1+2 m^{2}}\)

Case I:

\(\frac{1}{3}=\frac{-m}{1+2 m^{2}}\)

⇒ 1 + 2 m² = – 3m
⇒ 2m + 3m + 1 = 0
⇒ 2m² + 2m + m + 1 = 0
⇒ (m + 1) (2m + 1) = 0
⇒ m = – 1 or m = – \(\frac{1}{2}\)
If m = – 1, then the slopes of the lines are – 1 and – 2.
If m = – \(\frac{1}{2}\), then the slopes of the lines are – \(\frac{1}{2}\) and – 1.

Case II:
\(\frac{1}{3}=\frac{m}{1+2 m^{2}}\)
⇒ 2m² + 1 = 3m
⇒ 2m² – 2m – m + 1 = 0
⇒ (m – 1) (2m – 1) = 0
⇒ m = 1 or m = \(\frac{1}{2}\)
If m = 1, then the slopes of the lines are 1 and 2.
If m = \(\frac{1}{2}\), then the slopes of the lines are \(\frac{1}{2}\) and 1.
Hence, the slopes of the lines are – 1 and – 2 or – \(\frac{1}{2}\) and – 1 or 1 and 2 or \(\frac{1}{2}\) and 1.

Question 12.
A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that k – y₁ = m (h – x₁).
Answer.
The slope of the line passing through (x₁, y₁) and (h, k) is \(\frac{k-y_{1}}{h-x_{1}}\).

It is given that the slope of the line is

∴ \(\frac{k-y_{1}}{h-x_{1}}\) = m
⇒ k – y₁ = m (h – x₁)
Hence, k – y₁ = m (h – x₁).

Question 13.
If three points (h, 0), (a, b) and (0, k) lie on a line, show that \(\frac{a}{\boldsymbol{h}}+\frac{b}{\boldsymbol{k}}\) = 1.
Answer.
If the points A (h, 0), B (a, b) and C (0, k) lie on a line, then
Slope of AB = Slope of BC
\(\frac{b-0}{a-h}=\frac{k-b}{0-a}\)

⇒ \(\frac{b}{a-h}=\frac{k-b}{-a}\)

⇒ – ab = (k – b) (a – h)
⇒ – ab = ka – kh – ab + bh
⇒ ka + bh = kh
On dividing both sides by kh, we obtain
\(\frac{k a}{k h}+\frac{b h}{k h}=\frac{k h}{k h}\)

\(\frac{a}{h}+\frac{b}{k}\) = 1

Hence \(\frac{a}{h}+\frac{b}{k}\) = 1.

Question 14.
Consider the given population and year graph. Find the slope of the line AB and uaing it, find what will be the peculation in the year 2010?

Answer.
Since line AB passes through points A(1985, 92) and B(1995, 97), its slope is = \(\frac{97-92}{1995-1985}=\frac{5}{10}=\frac{1}{2}\)

Let y be the population in the year 2010.
Then, according to the given graph, line AB must pass through point C (2010, y)
∴ Slope of AB = Slope of BC

⇒ \(\frac{1}{2}=\frac{y-97}{2010-1995}\)

⇒ \(\frac{1}{2}=\frac{y-97}{15}\)

⇒ \(\frac{15}{2}\) = y – 97
⇒ y – 97 = 7.5
⇒ y = 7.5 + 97 = 104.5
Thus, the slope of line AB is \(\frac{1}{2}\), while in the year 2010, the population will be 104.5 crores.