NCERT Solutions Class 11 Mathematics Straight Lines Exercise 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 = -x/7 + 0  ………..1
This equation is of the form y = mx + c, where m = 1/7 and c = 0
Therefore, equation 1 is in the slope-intercept form, where the slope and the y-intercept are
-1/7 and 0 respectively.
(ii) The given equation is 6x + 3y – 5 = 0.
It can be written as
      3y = -6x + 5   ⇒ y = -6x/3 + 5/3
⇒ y = -2x + 5/3  ………….2
This equation is of the form y = mx + c, where m = -2 and c = 5/3
Therefore, equation 2 is in the slope-intercept form, where the slope and the y-intercept are
-2 and 5/3 respectively.
(iii) The given equation is y = 0.
It can be written as
y = 0 * x + 0   ……….3
This equation is of the form y = mx + c, where m = 0 and c = 0.
Therefore, equation 3 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
⇒ 3x/12 + 2y/12 = 1
⇒ x/4 + y/6 = 1 …………1
This equation is in the form x/a + y/b = 1
Where a = 4 and b = 6
Therefore, equation 1 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
      4x/6 – 3y/6 = 1
⇒ 2x/3 – y/2 = 1
⇒ x/(3/2) + y/(-2) = 1   …………..2
This equation is of the form x/a + y/b = 1, where a = 3/2 and b = –2.
Therefore, equation 2 is in the intercept form, where the intercepts on the x and y axes are
3/2 and –2 respectively.
(iii) The given equation is 3y + 2 = 0
It can be written as
      3y = -2
⇒ 3y/(-2) = 1
⇒ y/(-2/3) = 1       …………..3
The equation is of the form x/a + y/b, where a = 0 and b = -2/3
Therefore, equation 3 is in the intercept form, where the intercept on the y-axis is -2/3 and it
has no intercept on the x-axis.

Question 3:
Reduce the following equations into normal form. Find their perpendicular distances from the origin and
angle between perpendicular and the positive x-axis.
(i) x – √3y + 8 = 0                        (ii) y – 2 = 0                     (iii) x – y = 4
Answer :
(i) The given equation is x – √3y + 8 = 0                       
It can be reduced as:
      x – √3y = -8
⇒ -x + √3y = 8                        
On dividing both sides by √{(-1)² +(√3)²} = √(1 + 3) = √4 = 2, we get
⇒ -x/2 + √3y/2 = 8/2
⇒ -x/2 + √3y/2 = 4
⇒ (-1/2)x + (√3/2)y = 4
⇒ x cos 120° + y sin 120° = 4  ………….1
Equation 1 is in the normal form.
On comparing equation 1 with the normal form of equation of line x cos ω + y sin ω = p, we get
ω = 120° and p = 4
Thus, the perpendicular distance of the line from the origin is 4, while the angle between the
perpendicular and the positive x-axis is 120°.
(ii) The given equation is y – 2 = 0.
It can be reduced as 0.x + 1.y = 2
On dividing both sides by √{0² + 1²} = 1, we get
      0 * x + 1 * y = 2
⇒ x cos 90° + y sin 90° = 2  ………..1
Equation 1 is in the normal form.
On comparing equation 1 with the normal form of equation of line x cos ω + y sin ω = p, we get
ω = 90° and p = 2
Thus, the perpendicular distance of the line from the origin is 2, while the angle between the
perpendicular and the positive x-axis is 90°.
(iii) The given equation is x – y = 4.
It can be reduced as 1 * x + (–1)y = 4
On dividing both sides by √{1² + (-1)²} = √2, we get
     (1/√2)x + (-1/√2) = 4/√2
⇒ x cos (2π – π/4) + y sin (2π – π/4) = 2√2
⇒ x cos 315° + y sin 315° = 2√2  ………..1
Equation (1) is in the normal form.
On comparing equation 1 with the normal form of equation of line x cos ω + y sin ω = p, we get
ω = 315° and p = 2√2
Thus, the perpendicular distance of the line from the origin is 2√2, while the angle between
the perpendicular and the positive x-axis is 315°. 

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    ……..1
On comparing equation 1 with general equation of line Ax + By + C = 0, we get
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 = |Ax₁ + By₁ + C|/√(A² + B²)
The given point is (x₁, y₁) = (–1, 1)
Therefore, the distance of point (–1, 1) from the given line
= |12 * (-1) + (-5) * 1 + 82|/√{(12)² + (-5)²}
= |-12 – 5 + 82|/√ (144 + 25)
= |65|/√169
= 65/13
= 5 units

Question 5:
Find the points on the x-axis, whose distances from the line x/3 + y/4 = 1 are 4 units.

Answer:
The given equation of line is x/3 + y/4 = 1
⇒ (4x + 3y)/12 = 1
⇒ 4x + 3y = 12   ……………1
On comparing equation 1 with general equation of line Ax + By + C = 0, we get
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 = |Ax₁ + By₁ + C|/√(A² + B²)
Therefore,
      4 = |4a + 3 * 0 – 12|/√{4² + 3²}
⇒ 4 = |4a – 12|/√ (16 + 9)
⇒ 4 = |4a – 12|/5
⇒ |4a – 12| = 20
⇒ 4a – 12 = 20 or –(4a – 12) = 20
⇒ 4a = 20 + 12 or 4a = -20 + 12
⇒ 4a = 32 or 4a = -8
⇒ a = 8 or a = -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 = |C₁ – C₂|/√(A² + B²)

(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 parallel lines is
d = |C₁ – C₂|/√(A² + B²) = |-34 – 31|/√(15² + 8²) = |-65|/√ (225 + 64) = 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 = |C₁ – C₂|/√(A² + B²) = |p + r|/√(l² + l²) = |p + r|/√(2l²) = |p + r|/l√2 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
⇒ 4y = 3x + 2
⇒ y = 3x/4 + 2/4
⇒ y = 3x/4 + 1/2
which is of the form y = mx + c
So, slope of the given line = 3/4
It is known that parallel lines have the same slope.
Slope of the other line = m = 3/4
Now, the equation of the line that has a slope of 3/4 and passes through the point (–2, 3) is
      (y – 3) = 3{x – (-2)}/4
⇒ y – 3 = 3(x + 2)/4
⇒ 4(y – 3) = 3(x + 2)
⇒ 4y – 12 = 3x + 6
⇒ 3x – 4y + 18 = 0

Question 8:
Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

Answer:
The given equation of line is x – 7y + 5 = 0
⇒ 7y = x + 5
⇒ y = x/7 + 5/7
which is of the form y = mx + c
So, the slope of the given line = 1/7
The slope of the line perpendicular to the line having a slope of 1/7 is
m = -1/(1/7) = -7
The equation of the line with slope –7 and x-intercept 3 is given by
     y = m (x – d)
⇒ y = –7(x – 3)
⇒ y = –7x + 21
⇒ 7x + y = 21

Question 9:
Find angles between the lines √3x + y = 1 and x + √3y = 1

Answer:
The given lines are √3x + y = 1 and x + √3y = 1
⇒ y = -√3x + 1  ………..1
and y = -x/√3 + 1/√3   ……..2
The slope of the 1 is m₁ = -√3 and the slope of the line 2 is m₂ = -1/√3
The acute angle i.e., θ between the two lines is given by
      tan θ = |(m₁ – m₂)/(1 + m₁ * m₂)|
⇒ tan θ = |(-√3 + 1/√3)/{1 + (-√3) * (-1/√3)}|
⇒ tan θ = |(-√3 + 1/√3)/(1 + 1)|
⇒ tan θ = |{(-3 + 1)/√3}/2|
⇒ tan θ = |-2/2√3|
⇒ tan θ = |-1/√3|
⇒ tan θ = 1/√3
⇒ tan θ = tan 30°
⇒ θ = 30°
Thus, the angle between the given lines is either 30° or 180° – 30° = 150°.

Question 10:
The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0, at right angle. Find the value of h.

Answer:
The slope of the line passing through points (h, 3) and (4, 1) is
m₁ = (1 – 3)/(4 – h) = -2/(4 – h)
Given line is 7x – 9y – 19 = 0
⇒ 9y = 7x – 19
⇒ y = 7x/9 – 19/7
The slope of line 7x – 9y – 19 = 0 is m₂ = 7/9
It is given that the two lines are perpendicular.
So, m₁ * m₂ = -1
⇒ {-2/(4 – h)} * (7/9) = -1
⇒ -14/(36 – 9h) = -1
⇒ 36 – 9h = 14
⇒ 9h = 36 – 14
⇒ 9h = 22
⇒ h = 22/9
Thus, the value of h is 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 –x₁) + B (y – y₁) = 0.

Answer:
The slope of line Ax + By + C = 0 or
It is known that parallel lines have the same slope.
So, the slope of the other line = m = -A/B
The equation of the line passing through point (x₁, y₁) and having a slope –A/B is
      y – y₁ = m(x – x₁)
⇒ y – y₁ = (-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 angle of 60°. If slope of one line is 2, find equation of the other line.

Answer:
It is given that the slope of the first line, m₁ = 2
Let the slope of the other line be m₂
The angle between the two lines is 60°.
      tan θ = |(m₁ – m₂)/(1 + m₁ * m₂)|
⇒ tan 60° = |(2 – m₂)/(1 + 2m₂)|
⇒ √3 = ±(2 – m₂)/(1 + 2m₂)|
⇒ √3 = (2 – m₂)/(1 + 2m₂) or √3 = -(2 – m₂)/(1 + 2m₂)
⇒ √3(1 + 2m₂) = (2 – m₂) or √3(1 + 2m₂) = -(2 – m₂)
⇒ √3 + 2√3m₂ = 2 – m₂ or √3 + 2√3m₂ = -2 + m₂
⇒ √3 + 2√3m₂ + m₂ = 2 or √3 + 2√3m₂ – m₂ = -2
⇒ (2√3 + 1)m₂ = 2 – √3 or (2√3 – 1)m₂ = -2 – √3
⇒ m₂ = (2 – √3)/ (2√3 + 1) or m₂ = -(2 + √3)/(2√3 – 1)
Case 1: When m₂ = (2 – √3)/(2√3 + 1)
The equation of the line passing through point (2, 3) and having a slope of
m₂ = (2 – √3)/(2√3 + 1) is given by
      (y – 3) = {(2 – √3)/ (2√3 + 1)}(x – 2)
⇒ (y – 3)(2√3 + 1) = (2 – √3)(x – 2)
⇒ (2√3 + 1)y – 3(2√3 + 1) = (2 – √3)x – 2(2 – √3)
⇒ (√3 – 2)x + (2√3 + 1)y = -4 + 2√3 + 6√3 + 3
⇒ (√3 – 2)x + (2√3 + 1)y = -1 + 8√3
In this case, the equation of the other line is (√3 – 2)x + (2√3 + 1)y = -1 + 8√3
Case 2: When m₂ = -(2 + √3)/ (2√3 – 1)
The equation of the line passing through point (2, 3) and having a slope of
m₂ = -(2 + √3)/(2√3 + 1) is given by
      (y – 3) = {-(2 + √3)/ (2√3 – 1)}(x – 2)
⇒ (y – 3)(2√3 – 1) = -(2 + √3)(x – 2)
⇒ (2√3 – 1)y – 3(2√3 – 1) = -(2 + √3)x + 2(2 + √3)
⇒ (2 + √3)x + (2√3 – 1)y = 4 + 2√3 + 6√3 – 3
⇒ (2 + √3)x + (2√3 + 1)y = 1 + 8√3
In this case, the equation of the other line is (2 + √3)x + (2√3 + 1)y = 1 + 8√3

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).
Mid-point of AB = {(3 – 1)/2, (4 + 2)/2} = (1, 3)
Slope of AB = (2 – 4)/(-1 – 3) = -2/-4 = 1/2
So, the slope of the line perpendicular to AB = -1/(1/2) = -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 equation 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    3x – 4y – 16 = 0.

Answer: