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
