Inter 1st Year Maths Straight Lines Solutions Exercise 9d

Practicing the AP Board Solutions Class 11 Maths and Chapter 9 Inter 1st Year Maths Straight Lines Solutions Exercise 9d Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Straight Lines Solutions Exercise 9d

Straight Lines Exercise 9d Solutions

Straight Lines Class 11 Exercise 9d Solutions – Straight Lines 9d Exercise Solutions

I.

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.
Solution:
It is given that
(k – 3)x – (4 – k²)y + k² – 7k + 6 = 0 ………(1)
(a) Here, if the line is parallel to the x-axis slope of the line = slope of the x-axis
It can be written as
(4 – k²)y = (k – 3)x + k² – 7k + 6 = 0
we get 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
Here the slope of the given line = \(\frac{(k-3)}{\left(4-k^2\right)}\)
Consider the slope of the X-axis = 0
\(\frac{(k-3)}{\left(4-k^2\right)}\) = 0
By further calculation
k – 3 = 0
k = 3
Hence, if the given line is parallel to the x-axis, then the value of k is 3.

(b) Here, if the line is parallel to the y-axis, it is vertical, and the slope will be undefined.
So, the slope of the given line = \(\frac{(k-3)}{\left(4-k^2\right)}\)
Here, \(\frac{(k-3)}{\left(4-k^2\right)}\) is undefined at k² = 4
⇒ k² = 4
⇒ k = ±2
Hence, if the given line is parallel to the y-axis, then the value of k is ±2.

(c) Here, if the line passes through (0, 0), which is the origin satisfies the given equation of the line.
(k – 3) (0) – (4 – k²) (0) + k² – 7k + 6 = 0
By further calculation,
k² – 7k + 6 = 0
Separating the terms,
k² – 6k – k + 6 = 0
We get (k – 6) (k – 1) = 0
k = 1 (or) 6
Hence, if the given line passes through the origin, then the value of k is either 1 or 6.

Question 2.
Find the equations of the lines that cut off intercepts on the axes whose sum and product are 1 and -6, respectively.
Solution:
Consider the intercepts cut by the given lines on the a and b axes.
a + b = 1 ………(1)
ab = -6 ……..(2)
By solving both equations, we get
a = 3 and b = -2 or a = -2 and b = 3
We know that the equation of the line whose intercepts on the a and b axes is \(\frac{x}{a}+\frac{y}{b}\) = 1 or bx + ay – ab = 0
Case (i): a = 3, b = -2
So, the equation of the line is -2x + 3y + 6 = 0
i.e., 2x – 3y = 6
Case (ii): a = -2 and b = 3
So, the equation of the line is 3x – 2y + 6 = 0
i.e., -3x + 2y = 6
Hence, the required equation of the lines are 2x – 3y = 6 and -3x + 2y = 6

Question 3.
What are the points on the y-axis whose distance from the line \(\frac{x}{3}+\frac{y}{4}\) = 1 is 4 units.
Solution:
Consider (0, b) as the point on the y-axis whose distance from the line \(\frac{x}{3}+\frac{y}{4}\) = 1 is 4 units.
It can be written as 4x + 3y – 12 = 0 …….(1)
By comparing equation (1) to the general equation of a line, Ax + By + C = 0, from line (x₁, y₁) is written as
d = \(\frac{\left|\mathrm{Ax}_1+\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\)
If (0, b) is the point on the y-axis whose distance from \(\frac{x}{3}+\frac{y}{4}\) = 1 is 4 units, then
4 = \(\frac{|4(0)+3(b)-12|}{\sqrt{4^2+3^2}}\)
By further calculation
4 = \(\frac{|3 b-12|}{5}\)
20 = |3b – 12|
We get 20 = ±(3b – 12)
Here, 20 = (3b – 12) or 20 = -(3b – 12)
It can be written as 3b = 20 + 12 or 3b = -20 + 12
So, we get b = \(\frac {32}{3}\) or b = \(\frac {-8}{3}\)
Hence, the required points (0, \(\frac {32}{3}\)) and (0, \(\frac {-8}{3}\))

Question 4.
Find the perpendicular distance from the origin to the line joining the points (cos θ, sin θ) and (cos φ, sin φ).
Solution:
Hence, the equation of the line joining the points (cos θ, sin θ) and (cos φ, sin φ) is written as
y – sin θ = \(\frac{\sin \phi-\sin \theta}{\cos \phi-\cos \theta}\) (x – cos θ)
By cross multiplication
y(cos φ – cos θ) – sin θ (cos φ – cos θ) = x(sin φ – sin θ) – cos θ (sin φ – sin θ)
By multiplying the terms, we get
x(sin θ – sin φ) + y(cos φ – cos θ) + cos θ sin φ – cos θ sin θ – sin θ cos φ + sin θ cos θ = 0
on further simplification
x(sin θ – sin φ) + y(cos φ – cos θ) + sin(φ – θ) = 0
So we get Ax + By + C = 0,
where A = sin θ – sin φ, B = cos φ – cos θ, and C = sin (φ – θ)
We know that the perpendicular distance (d) of a line Ax + By + C = 0 from (x₁, y₁) is written as
d = \(\frac{\left|\mathrm{Ax}_1+\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\)
So, the perpendicular distance (d) of the given line from (x₁, y₁) = (0, 0) is

Question 5.
Find the equation of the line parallel to the y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.
Solution:
Here, the equation of any line parallel to the y-axis of the form
x = a …………(1)
Two given lines are
x – 7y + 5 = 0 ……….(2)
3x + y = 0 ………..(3)
By solving equations (2) and (3),
We get x = \(\frac {-5}{22}\) and y = \(\frac {15}{22}\)
\(\left(\frac{-5}{22}, \frac{15}{22}\right)\) is the point of intersection of lines (2) and (3)
Hence, the required point through the line x = \(\frac {-5}{22}\)

Question 6.
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.
Solution:
It is given that \(\frac{x}{4}+\frac{y}{6}\) = 1
We can write it as 3x + 2y – 12 = 0
So, we get y = \(\frac {-3}{2}\)x + 6, which is of the from y = mx + c
Here, the slope of the given line = \(\frac {-3}{2}\)
So, the slope of line perpendicular to the give line = \(\frac{-1}{\left(\frac{-3}{2}\right)}=\frac{2}{3}\)
Consider the given line intersects the y-axis at (0, y)
By substituting x = 0 in the equation of the given line,
\(\frac {y}{6}\) = 1
y = 6
Hence, the given line intersects the y-axis at (0, 6).
We know that the equation of the line that has a slope of \(\frac {2}{3}\) and passes through the point (0, 6) is
(y – 6) = \(\frac {2}{3}\)(x – 0)
By further calculation,
3y -18 = 2x
So, we get 2x – 3y + 18 = 0
Hence, the required equation of the line is 2x – 3y + 18 = 0

Question 7.
Find the length of the perpendicular drawn from the point (3, 4) to the straight line 3x – 4y + 10 = 0.
Solution:
Given lines 3x – 4y + 10 = 0 and point (3, 4)
The length of the perpendicular distance from the line is
\(\left|\frac{3(3)-4(4)+10}{\sqrt{9+16}}\right|=\frac{|9-16+10|}{\sqrt{25}}\) = \(\frac {3}{5}\)

Question 8.
Find the distance between the following parallel lines.
(i) 3x – 4y = 12, 3x – 4y = 7
(ii) 5x – 3y – 4 = 0, 10x – 6y – 9 = 0
Solution:
(i) Given lines are 3x – 4y – 12 = 0, 3x – 4y – 7 = 0
Distance between theparallel lines = \(\left|\frac{-12+7}{\sqrt{9+6}}\right|=\frac{5}{5}\) = 1

(ii) Given lines are 5x – 3y – 4 = 0, 10x – 6y – 8 = 0
⇒ 10x – 6y – 8 = 0, 10x – 6y – 9 = 0
Distance between the parallel lines = \(\frac{|-8+9|}{\sqrt{100+36}}=\frac{1}{\sqrt{136}}=\frac{1}{2 \sqrt{34}}\)

Question 9.
Find the value of k if the straight lines y – 3kx + 4 = 0 and (2k – 1)x – (8x – 1)y – 6 = 0 are perpendicular.
Solution:
Given lines are y – 3kx + 4 = 0, (2k – 1)x – (8k – 1)y – 6 = 0
⇒ 3xk – y – 4 = 0, (2k – 1)x – (8k – 1)y – 6 = 0
Given lines are perpendicular
⇒ 3k(2k – 1) + ((-1) (-8k – 1)) = 0
⇒ 6k² – 3k + 8k – 1 = 0
⇒ 6k² + 5k – 1 = 0
⇒ 6k² + 6k – k – 1 = 0
⇒ (6k – 1) (k + 1) = 0
⇒ k = -1, \(\frac {1}{6}\)

Question 10.
(-4, 5) is a Vertex of a square, and one of its diagonals is 7x – y + 8 = 0. Find the equation of the other diagonal.
Solution:
The other diagonal is the line through (-4, 5) and perpendicular to 7x – y + 8 = 0
Its equation is 1(x + 4) + 7(y – 5) = 0
⇒ x + 7y – 31 = 0

II.

Question 1.
Find the area of the triangle formed by the lines y – x = 0, x + y = 0, and x – k = 0.
Solution:
It is given that
-1 – x = 0 ……….(1)
x + y = 0 ………(2)
x – k = 0 ……….(3)
Here, the point of intersection of lines (1) & (2) is x = 0 and y = 0
Lines (2) and (3) is x = k and y = -k
lines (3) and (1) are x = k and y = k
So, the vertices of the triangles formed by the three given lines are (0, 0), (k, -k), and (k, k).
Here, the area triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) is
\(\frac {1}{2}\) |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
So, the area of the triangle formed by the three given lines = \(\frac {1}{2}\) |0(-k – k) + k(k – 0) + k(0 + k)| square units
= \(\frac {1}{2}\) |k² + k²| square units.
= \(\frac {1}{2}\) |2k²|
= k² square units.

Question 2.
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.
Solution:
3x + y – 2 = 0 ……….(1)
px + 2y – 3 = 0 ………(2)
2x – y – 3 = 0 ……….(3)
By solving equations (1) and (3),
We get x = 1 and y = -1
Here, the three lines intersect at one point, and the point of intersection of lines (1) and (3) will also satisfy line (2)
p(1) + 2(-1) – 3 = 0
By further calculation,
p – 2 – 3 = 0
So we get p = 5
Hence, the required value of p is 5.

Question 3.
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.
Solution:
It is given that
y = m₁x + c₁ ………..(1)
y = m₂x + c₂ ……….(2)
y = m₃x + c₃ ………..(3)
By subtracting equation (1) from (2), we get
0 = (m₂ – m₁) x + (c₂ – c₁)
(m₁ – m₂)x = c₂ – c₁
So, we get x = \(\frac{c_2-c_1}{m_1-m_2}\)
By substituting this value in equation (1), we get

Here \(\left(\frac{c_2-c_1}{m_1-m_2}, \frac{m_1 c_2-m_2 c_1}{m_1-m_2}\right)\) is the point of intersection of lines (1) and (2)
Lines (1), (2), and (3) are concurrent,
So the point of intersection of lines (1) and (2) will satisfy equation (3)
\(\frac{m_1 c_2-m_2 c_1}{m_1-m_2}=m_3\left(\frac{c_2-c_1}{m_1-m_2}\right)+c_3\)
By multiplying the terms and taking the LCM
\(\frac{m_1 c_2-m_2 c_1}{m_1-m_2}\) = \(\frac{m_3 c_2-m_3 c_1+c_3 m_1-c_3 m_2}{m_1-m_2}\)
By cross multiplication
m₁c₂ – m₂c₁ – m₃c₂ + m₃c₁ – c₃m₁ + c₃m₂ = 0
taking out the common terms,
m₁(c₂ – c₃) + m₂(c₃ – c₁) + m₃(c₁ – c₂) = 0
Therefore, m₁(c₂ – c₃) + m₂(c₃ – c₁) + m₃(c₁ – c₂) = 0

Question 4.
Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x – 2y = 3.
Solution:
Consider m₁ as the slope of the required line
It can be written as \(\frac{1}{2} x-\frac{3}{2}\) which is of the form y = mx + c
So, the slope of the given line m₂ = \(\frac {1}{2}\)
We know that the angle between the required line and the line x – 2y = 3 is 45°.
If θ is the acute angle between lines l₁ and l₂ with slopes m₁ and m₂


Case(i): m₁ = 3
Here, the equation of the line passing through (3, 2) and having a slope of 3 is y – 2 = 3(x – 3)
By further calculation,
y – 2 = 3x – 9
So, we get
3x – y = 7
Case (ii): m₁ = \(\frac {-1}{3}\)
Here, the equation of the line passing through (3, 2) and having slope \(\frac {-1}{3}\) is
y – 2 = \(\frac {-1}{3}\)(x – 3)
By further calculation,
3y – 6 = -x + 3
So, we get x + 3y = 9
Hence, the equations of the line are 3x – y = 7 and x + 3y = 9.

Question 5.
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.
Solution:
Consider the equation of the line having equal intercepts on the axes as \(\frac{x}{a}+\frac{y}{a}\) = 1
It can be written as
x + y = a ……….(1)
By solving equations 4x + 7y – 3 = 0 and 2x – 3y +1 – 0
we get x = \(\frac {1}{13}\) and y = \(\frac {5}{13}\)
\(\left(\frac{1}{13}, \frac{5}{13}\right)\) is the point of intersection of two given lines.
We know that equation (1) passes through the point \(\left(\frac{1}{13}, \frac{5}{13}\right)\)
\(\frac{1}{13}+\frac{5}{13}\) = a
a = \(\frac {6}{13}\)
Here, equation (1) becomes
x + y = \(\frac {6}{13}\)
13x + 13y = 6
Hence, the required equation of the line is 13x + 13y = 6.

Question 6.
Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is \(\frac{y}{x}=\frac{m \pm \tan \theta}{1 \mp m \tan \theta}\).
Solution:
Consider y = m₁x as the equation of the line passing through the origin.
It is given that the line makes an angle θ with the line y = mx + c, then angle θ is written as

We can write it as
tan θ + \(\frac {y}{x}\) m tan θ = \(\frac {y}{x}\) – m
By further simplification
m tan θ = \(\frac {y}{x}\) (1 – m tan θ)
So, we get \(\frac{y}{x}=\frac{m+\tan \theta}{1-m \tan \theta}\)

Case (ii):
tan θ = \(-\left(\frac{\frac{y}{x}-m}{1+\frac{y}{x} m}\right)\)
We can write it as
tan θ + \(\frac {y}{x}\) m tan θ = –\(\frac {y}{x}\) + m
By further simplification
\(\frac {y}{x}\) (1 + m tan θ) = m – tan θ
So we get \(\frac{y}{x}=\frac{m+\tan \theta}{1-m \tan \theta}\)

Case (iii):
tan θ = \(-\left(\frac{\frac{\mathrm{y}}{\mathrm{x}}-\mathrm{m}}{1+\frac{\mathrm{y}}{\dot{\mathrm{x}}} \mathrm{~m}}\right)\)
We can write it as
tan θ + \(\frac {y}{x}\) m tan θ = –\(\frac {y}{x}\) + m
By further simplification
\(\frac {y}{x}\) (1 + m tan θ) = m – tan θ
So we get \(\frac{\mathrm{m} \pm \tan \theta}{1 \pm \mathrm{m} \tan \theta}\)

Question 7.
In what ratio, the line segment joining (-1, 1) and (5, 7) divided by the line x + y = 4?
Solution:
We know that 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)
By further calculation
y – 1 = x + 1
So, we get x – y + 2 = 0
Hence, the line joining the points (-1, 1) and (5, 7) is divided by the line x + y = 4 in the ratio 1 : 2.

Question 8.
Find the equations of the straight lines passing through (1, 3) and (i) parallel to (ii) perpendicular to the line passing through the points (3, -5) and (-6, 1).
Solution:
Let A (3, -5), B(-6, 1)
Slope of \(\overline{\mathrm{AB}}=\frac{1+5}{-6-3}=\frac{6}{-9}=\frac{-2}{3}\)
Slope of the line perpendicular to \(\overline{\mathrm{AB}}=\frac{3}{2}\)
(i) Equation of the line through (1, 3) and parallel to \(\overline{\mathrm{AB}}\) is
y – 3 = \(\frac {-2}{3}\) (x – 1)
⇒ 3y – 9 = -2x + 2
⇒ 2x + 3y – 11 = 0

(ii) Equation of the line through (1, 3) and perpendicular to \(\overline{\mathrm{AB}}\) is
y – 3 = \(\frac {3}{2}\)(x – 1)
⇒ 2y – 6 = 3x – 3
⇒ 3x – 2y + 3 = 0

Question 9.
Find the equation of the line perpendicular to the line 3x + 4y + 6=0 and making an intercept of -4 on the X-axis.
Solution:
x-intercept of the line = -4
⇒ line passes through (-4, 0)
Equation of the required line is 4(x + 4) – 3(y – 0) = 0
⇒ 4x – 3y + 16 = 0

Question 10.
A(-1, 1), B(5, 3) are opposite vertices of a square in the XY plane. Find the equation of the other diagonal (not passing through A, B) of the square.
Solution:
Slope of \(\overline{\mathrm{AB}}\) is \(\frac{3-1}{5+1}=\frac{1}{3}\)
∴ Slope of the other diagonal is -3
Midpoint of AB is (2, 2)
equation to the other diagonal is y – 2 = -3(x – 2)
⇒ 3x + y – 8 = 0

Question 11.
Find the foot of the perpendicular drawn from (3, 0) upon the straight line 5x + 12y – 41 = 0.
Solution:
If (b, k) is the foot of the perpendicular from (3, 0) to the line 5x + 12y – 41 = 0, then

∴ Foot of the perpendicular are \(\left(\frac{68}{25}, \frac{49}{25}\right)\).

Question 12.
Show that the distance of the point (6, -2) from the line 4x + 3y = 12 is half the distance of the point (3, 4) from the line 4x – 3y = 12.
Solution:
Given P = (6, -2) and L₁ = 4x + 3y = 12
D₁ = Distance of the point to the line is \(\frac{|4(6)+3(-2)-12|}{\sqrt{16+9}}\)
D₁ = \(\frac {6}{5}\)
P₂ = (3, 4) and L₂ = 4x – 3y = 12
D₂ = Distance of the point to the line is \(\frac{\mid 4(3)+(-3)(4)-12) \mid}{\sqrt{16+9}}\)
D₂ = \(\frac {12}{5}\)
∴ D₁ = \(\frac {1}{2}\) D₂
⇒ \(\frac{6}{5}=\frac{1}{2}\left(\frac{12}{5}\right)\)

III.

Question 1.
Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.
Solution:
It is given that
2x – y = 0 ……….(1)
4x + 7y + 5 = 0 ………(2)
Here, A(1, 2) is a point on the line (1)
Consider B as the point of intersection of lines (1) and (2)
By solving equations (1) and (2), we get
x = \(\frac {-5}{81}\) and y = \(\frac {-5}{9}\)
So, the coordinates of point B are \(\left(\frac{-5}{18}, \frac{-5}{9}\right)\)
From the distance formula, the distance between A and B
AB = \(\sqrt{\left(1+\frac{5}{18}\right)^2+\left(2+\frac{5}{9}\right)^2}\) units
By taking LCM
x – y + 2 = 0 ……….(1)
So the equation of the given line is
x + y – 4 = 0 …………(2)
Here, the point of intersection of lines (1) and (2) is given by x = 1 and y = 3
Consider (1, 3) dividing the line segment joining (-1, 1) and (5, 7) in the ratio 1 : k
Using the section formula

Question 2.
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.
Solution:
Consider y = mx + c as the line passing through the point (-1, 2).
So we get 2 = m(-1) + c
By further calculation,
2 = -m + c
c = m + 2
Substituting the value of c
y = mx + m + 2 ………..(1)
So, the given line is
x + y = 4 ……….(2)
By solving both equations, we get
x = \(\frac{2-m}{m+1}\) and y = \(\frac{5 m+2}{m+1}\)
\(\left(\frac{2-m}{m+1}, \frac{5 m+2}{m+1}\right)\) is the point of intersection of lines (1) and (2)
Here, the point is at a distance of 3 units from (1, 2)
From the distance formula.

Question 3.
The hypotenuse of a right-angled triangle has its ends at the points (1, 3) and (-4, 1). Find an equation of the legs (perpendicular sides) of the triangle that are parallel to the axes.
Solution:
Consider ABC as the right-angled triangle where ∠C = 90°
Here, infinity such lines are present, m is the Slope of AC
So, the slope of BC = \(-\frac {1}{m}\)
Equation of AC
y – 3 = m(x – 1)
By cross multiplication,
x – 1 = \(\frac {1}{m}\)(y – 3)
Equation of BC
y – 1 = \(-\frac {1}{m}\) (x + 4)
By cross multiplication
x + 4 = -m(y – 1)
By considering the values of m,
We get m = 0
So, we get y – 3 = 0, x + 4 = 0
If m = ∞,
So, we get , x – 1 = 0, y – 1 = 0
we get x = 1, y = 1
\(\sqrt{\left(\frac{2-m}{m+1}+1\right)^2+\left(\frac{5 m+2}{m+1}-2\right)^2}\) = 3
Squaring on both sides
\(\left(\frac{2-m+m+1}{m+1}\right)^2\) + \(\left(\frac{5 m+2-2 m-2}{m+1}\right)^2\) = 3²
By further calculation
\(\frac{9}{(m+1)^2}+\frac{9 m^2}{(m+1)^2}\)
Dividing the equation by 9
\(\frac{1+m^2}{(m+1)^2}\) = 1
By cross multiplication,
1 + m² = m² + 1 + 2m
So, we get 2m = 0
⇒ m = 0
Hence, the slope of the required line must be zero.
i.e., the line must be parallel to the x-axis.

Question 4.
Find the image of the given points with respect to the given straight line.
(i) (3, 8)…. x + 3y = 7
(ii) (1, 2)…. 3x + 4y – 1 = 0
Solution:
x + 3y = 7 ………..(1)
Consider B (a, b) as the image of point A(3, 8)
So line (1) is the perpendicular bisector of AB
Here, Slope of AB = \(\frac{b-8}{a-3}\)
Slope of line (1) = \(\frac {-1}{3}\)
Line (1) is perpendicular to AB
\(\left(\frac{b-8}{a-3}\right) \times\left(-\frac{1}{3}\right)\) = -1
By further calculation
\(\frac{h-8}{3 a-9}\) = 1
By cross multiplication
b – 8 = 3a – 9
3a – b = 1 ………..(2)
We know that
mid point of AB = \(\left(\frac{a+3}{2}, \frac{b+8}{2}\right)\)
So the midpoint of line segment AB will satisfy line (1)
From equation (1)
\(\left(\frac{a+3}{2}\right)+3\left(\frac{b+8}{2}\right)\) = 7
By further calculation
a + 3 + 3b + 24 = 14
on further simplification,
a + 3b = -13 ………(3)
By solving equations (2) and (3)
we get a = -1 and b = -4
Hence, the image of the given point with respect to the given line is (-1, 4).

Question 5.
If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.
Solution:
It is given that
y = 3x + 1 ……….(1)
2y = x + 3 ……….(2)
y = mx + 4 …………(3)
Here, the slopes of
Line (1), m₁ = 3
Line (2), m₂ = \(\frac {1}{2}\)
Line (3), m₃ = m
We know that lines (1) and (2) are equally inclined to line (3), which means that the angle between lines (1) and (3) equals the angle between lines (2) and (3).
\(\left|\frac{m_1-m_3}{1+m_1 m_3}\right|=\left|\frac{m_2-m_3}{1+m_2 m_3}\right|\)
Substituting the values, we get

By cross multiplication
(3 – m) (m + 2) = (1 – 2m) (1 + 3m)
on further calculation,
-m² + m + 6 = 1 + m – 6m²
So, we get 5m² + 5 = 0
Dividing the equation by 5,
m² + 1 = 0
m = \(\sqrt{-1}\), which is not real.
Therefore, this case is not possible
If \(\frac{3-m}{1+3 m}=-\left(\frac{1-2 m}{m+2}\right)\)
By cross multiplication
(3 – m) (m + 2) = -(1 – 2m) (1 + 3m)
On further calculation
-m² + m + 6 = -(1 + m – 6m²)
So we get 7m² – 2m – 7 = 0
Here we get
m = \(\frac{2 \pm \sqrt{4-4(7)(-7)}}{2(7)}\)
By further simplification
m = \(\frac{2 \pm 2 \sqrt{1+49}}{14}\)
We can write it as m = \(\frac{1 \pm 5 \sqrt{2}}{7}\)
Hence, the required value of m is \(\frac{1 \pm 5 \sqrt{2}}{7}\).

Question 6.
If the 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.
Solution:
It is given that
x + y – 5 = 0 ………(1)
3x – 2y + 7 = 0 ……….(2)
Here, the perpendicular distances of P(x, y) from lines (1) and (2) are written as

By further calculation
√13 |x + y – 5| + √2 |3x – 2y + 7| – 10√26 = 0
It can be written as √13 (x + y – 5) + √2 (3x – 2y + 7) – 10√26 = 0
Now by assuming (x + y – 5) and (3x – 2y + 7) are possitive.
√13x + √13y – 5√13 + 3√2x – 2√2y + 7√2 – 10√2b = 0
Taking out the common terms
x(√13 + 3√2) + y(√13 – 2√2) + (7√2 – 5√13 – 10√26) = 0
Which is the equation of a line,
In the same way, we can find the equation of the line for any signs of (x + y – 5) and (3x – 2y + 7).
Hence, point p must move on a line.

Question 7.
Find the equation of the line that is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
Solution:
It is given that
9x + 6y – 7 = 0 ……….(1)
3x + 2y + 6 = 0 ……….(2)
Consider p(h, k) to be the arbitrary point that is equidistant from lines (1) and (2).
Here the perpendicular distance of p(h, k) from line (1) is written as

Similarly the perpendicular distance of p(h, k) from line (2) is written as
d₂ = \(\frac{|3 h+2 k+6|}{\sqrt{(3)^2+(2)^2}}=\frac{|3 h+2 k+6|}{\sqrt{13}}\)
We know that p(h, k) is equidistant from lines (1) and (2)
d₁ = d₂, substituting the values
\(\frac{|9 h+6 k-7|}{3 \sqrt{13}}=\frac{|3 h+2 k+6|}{\sqrt{13}}\)
By further calculation
|9h + 6k – 7| = 3 |3h + 2k + 6|
It can be written as |9h + 6k – 7| = ±3(3h + 2k + 6)
Here, 9h + 6k – 7 = 3 (3h + 2k + 6) or 9h + 6k – 7 = -3(3h + 2k + 6)
9h + 6k – 7 = 3(3h + 2k + 6) is not possible as 9h + 6k – 7 = 3(3h + 2k + 6)
By further calculation, -7 = 18 (which is wrong)
We know that 9h + 6k – 7 = -3(3h + 2k + 6)
By multiplication,
9h + 6k – 7 = -9h – 6k – 18
We get 18h + 12k + 11 = 0
Hence, the required equation of the line is 18x + 12y + 11 = 0.

Question 8.
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.
Solution:
Consider the coordinates of point A as (a, 0)
Construct a line (AL) that is perpendicular to the x-axis

Here, the angle of incidence is equal to the angle of reflection
∠BAL = ∠CAL = φ
∠CAX = θ
It can be written as
∠OAB = 180° – (θ + 2φ) = 180° – [θ + 2(90° – θ)]
on further calculation,
= 180° – θ – 180° + 2θ
= θ
So, we get ∠BAX = 180° – θ
Slope of line AC = \(\frac{3-0}{5-a}\)
tan θ = \(\frac{3}{5-a}\) ……….(1)
Slope of line AB = \(\frac{2-0}{1-a}\)
We get tan(180° – θ) = \(\frac{2}{1-a}\)
By further calculation
-tan θ = \(\frac{2}{1-a}\)
tan θ = \(\frac{2}{a-1}\) …….(1)
From equations (1) and (2) we get
\(\frac{3}{5-a}=\frac{2}{a-1}\)
By cross multiplication,
3a – 3 = 10 – 2a
We get a = \(\frac {13}{5}\)
Hence the coordinates of point A are (\(\frac {13}{5}\), 0).

Question 9.
Prove that the product of the lengths of the perpendiculars drawn from the points \(\left(\sqrt{a^2-b^2}, 0\right)\) and \(\left(-\sqrt{a^2-b^2}, 0\right)\) to the line \(\frac {x}{a}\) cos θ + \(\frac {y}{b}\) sin θ = 1 is b².
Solution:
It is given that
\(\frac {x}{a}\) cos θ + \(\frac {y}{b}\) sin θ = 1
We can write it as
bx cos θ + ay sin θ – ab = 0 ………..(1)
Here, the length of the perpendicular from point \(\left(\sqrt{a^2-b^2}, 0\right)\) to line (1)

Question 10.
A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.
Solution:
It is given that
2x – 3y + 4 = 0 …….(1)
3x + 4y – 5 = 0 ………(2)
6x – 7y + 8 = 0 ………(3)
Here, the person is standing at the junction of the paths represented by lines (1) and (2).
By solving equations (1) and (2)
We get x = \(\frac {-1}{17}\) and y = \(\frac {22}{17}\)
Hence, the person is standing at point \(\left(\frac{-1}{17}, \frac{22}{17}\right)\)
We know that the person can reach path (3) in the least time if they walk along the perpendicular line to (3) from point \(\left(\frac{-1}{17}, \frac{22}{17}\right)\)
Here, the slope of line (3) = \(\frac {6}{7}\)
We get the slope of the line perpendicular to the line (3) = \(\frac{-1}{\left(\frac{6}{7}\right)}=\frac{-7}{6}\)
So, the equation of the line passing through \(\left(\frac{-1}{17}, \frac{22}{17}\right)\) and having a slope of \(\frac {-7}{6}\) is written as
By expanding using a formula

Therefore, it is proved.
\(\left(y-\frac{22}{17}\right)=-\frac{7}{6}\left(x+\frac{1}{17}\right)\)
By further calculation,
6(17y – 22) = -7(17x + 1)
By multiplication,
102y – 132 = -119x – 7
we get 119x + 102y = 125
Therefore, the path that the person should follow is 119x + 102y = 125.

Question 11.
Show that the lines x – 7y – 22 = 0, 3x + 4y + 9 = 0 and 7x + y – 54 = 0 form a right angled isosceles triangle.
Solution:
Let x – 7y – 22 = 0 ………(1)
3x + 4y + 9 = 0 ………(2)
and 7x + y – 54 = 0 ……….(3)
Slope of (1) is \(\frac {1}{7}\)
Slope bf (2) is \(\frac {-3}{4}\)
slope of (3) is -7
If α is an angle between the lines (1) and (2), then
tan α = \(\frac{\left(\frac{1}{7}\right)+\left(\frac{3}{4}\right)}{1+\frac{1}{7}\left(\frac{-3}{3}\right)}=\frac{4+21}{28-3}\) = 1
⇒ α = \(\frac{\pi}{4}\)
∴ Angles between (1) and (2) are
⇒ α = \(\frac{\pi}{4}\), \(\frac{3\pi}{4}\)
If β is an angle between the lines (1) and (3), then
tan β = \(\frac{\frac{1}{7}+7}{1+\left(\frac{1}{7}\right)(-7)}\) not defined
∴ β = \(\frac{\pi}{2}\)
Since one angle of a triangle is \(\frac{\pi}{2}\), the other two angles are acute, and hence they are \(\frac{\pi}{4}\), \(\frac{\pi}{4}\)
∴ The given lines form an isosceles right-angled triangle.

Question 12.
Find the equations of the straight lines passing through the point (-3, 2) and making an angle of 45° with the straight line 3x – y + 4 = 0.
Solution:
Let m be the slope of the required line slope of the given line is 3.
Since the required line makes an angle of 45° with the given line
tan 45° = \(\left|\frac{m-2}{1+3 m}\right|\)
⇒ (1 + 3m)² = (m – 3)²
⇒ 1 + 9m² + 6m = m² + 9 – 6m
⇒ 8m² + 12m – 8 = 0
⇒ 2m² + 3m – 2 = 0
⇒ (m + 2) (2m – 1) = 0
⇒ m = -2 or m = \(\frac {1}{2}\)
If m = -2 then equation of the line is y – 2 = -2(x + 3)
⇒ y – 2 = -2x – 6
⇒ 2x + y + 4 = 0
If m = \(\frac {1}{2}\) then the equation of the line is
y – 2 = \(\frac {1}{2}\)(x + 3)
⇒ 2y – 4 = x + 3
⇒ x – 2y + 7 = 0