NCERT Solutions Class 11th Maths Chapter – 10 Straight Lines Exercise 10.3

NCERT Solutions Class 11th Maths Chapter – 10 Straight Lines Exercise 10.3

NCERT Solutions Class 11th Maths Chapter – 10 Straight Lines Exercise 10.3

?Chapter – 10?

✍Straight Lines✍

?Exercise 10.3?

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

?‍♂️solution – (i) x + 7y = 0
Given –
The equation is x + 7y = 0
Slope – intercept form is represented in the form ‘y = mx + c’, where m is the slope and c is the y intercept
So, the above equation can be expressed as
y = -1/7x + 0
∴ The above equation is of the form y = mx + c, where m = -1/7 and c = 0.

(ii) 6x + 3y – 5 = 0
Given –
The equation is 6x + 3y – 5 = 0
Slope – intercept form is represented in the form ‘y = mx + c’, where m is the slope and c is the y intercept
So, the above equation can be expressed as
3y = -6x + 5
y = -6/3x + 5/3
= -2x + 5/3
∴ The above equation is of the form y = mx + c, where m = -2 and c = 5/3.

(iii) y = 0
Given –
The equation is y = 0
Slope – intercept form is given by ‘y = mx + c’, where m is the slope and c is the y intercept
y = 0 × x + 0
∴ The above equation is of the form y = mx + c, where m = 0 and c = 0.

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

?‍♂️solution – (i) 3x + 2y – 12 = 0
Given –
The equation is 3x + 2y – 12 = 0
Equation of line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepts on x axis and y – axis respectively.
So, 3x + 2y = 12
now let us divide both sides by 12, we get
3x/12 + 2y/12 = 12/12
x/4 + y/6 = 1
∴ The above equation is of the form x/a + y/b = 1, where a = 4, b = 6
Intercept on x – axis is 4
Intercept on y – axis is 6

(ii) 4x – 3y = 6
Given –
The equation is 4x – 3y = 6
Equation of line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepts on x axis and y – axis respectively.
So, 4x – 3y = 6
Now let us divide both sides by 6, we get
4x/6 – 3y/6 = 6/6
2x/3 – y/2 = 1
x/(3/2) + y/(-2) = 1
∴ The above equation is of the form x/a + y/b = 1, where a = 3/2, b = -2
Intercept on x – axis is 3/2
Intercept on y – axis is -2

(iii) 3y + 2 = 0
Given –
The equation is 3y + 2 = 0
Equation of line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepts on x axis and y – axis respectively.
So, 3y = -2
Now, let us divide both sides by -2, we get
3y/-2 = -2/-2
3y/-2 = 1
y/(-2/3) = 1
∴ The above equation is of the form x/a + y/b = 1, where a = 0, b = -2/3
Intercept on x – axis is 0
Intercept on y – axis is -2/3

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

?‍♂️solution – (i) x – √3y + 8 = 0
Given –
The equation is x – √3y + 8 = 0
Equation of line in normal form is given by x cos θ + y sin θ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’ is perpendicular distance from origin.
So now, x – √3y + 8 = 0
x – √3y = -8
Divide both the sides by √(1² + (√3)²) = √(1 + 3) = √4 = 2
x/2 – √3y/2 = -8/2
(-1/2)x + √3/2y = 4
This is in the form of: x cos 120o + y sin 120o = 4
∴ The above equation is of the form x cos θ + y sin θ = p, where θ = 120° and p = 4.
Perpendicular distance of line from origin = 4
Angle between perpendicular and positive x – axis = 120°

(ii) y – 2 = 0
Given –
The equation is y – 2 = 0
Equation of line in normal form is given by x cos θ + y sin θ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’ is perpendicular distance from origin.
So now, 0 × x + 1 × y = 2
Divide both sides by √(0² + 1²) = √1 = 1
0 (x) + 1 (y) = 2
This is in the form of: x cos 90o + y sin 90o = 2
∴ The above equation is of the form x cos θ + y sin θ = p, where θ = 90° and p = 2.
Perpendicular distance of line from origin = 2
Angle between perpendicular and positive x – axis = 90°

(iii) x – y = 4
Given –
The equation is x – y + 4 = 0
Equation of line in normal form is given by x cos θ + y sin θ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’ is perpendicular distance from origin.
So now, x – y = 4
Divide both the sides by √(1² + 1²) = √(1+1) = √2
x/√2 – y/√2 = 4/√2
(1/√2)x + (-1/√2)y = 2√2
This is in the form: x cos 315o + y sin 315o = 2√2
∴ The above equation is of the form x cos θ + y sin θ = p, where θ = 315° and p = 2√2.
Perpendicular distance of line from origin = 2√2
Angle between perpendicular and positive x – axis = 315°

4. Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

?‍♂️solution – Given –
The equation of the line is 12(x + 6) = 5(y – 2).
12x + 72 = 5y – 10
12x – 5y + 82 = 0 … (1)
Now, compare equation (1) with general equation of line Ax + By + C = 0, where A = 12, B = –5, and C = 82
Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

∴ The distance is 5units.

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

?‍♂️solution – Given –
The equation of line is x/3 + y/4 = 1
4x + 3y = 12
4x + 3y – 12 = 0 …. (1)
Now, compare equation (1) with general equation of line Ax + By + C = 0, where 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.
So, the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

|4a – 12| = 4 × 5
± (4a – 12) = 20
4a – 12 = 20 or – (4a – 12) = 20
4a = 20 + 12 or 4a = -20 + 12
a = 32/4 or a = -8/4
a = 8 or a = -2
∴ The required points on the x – axis are (-2, 0) and (8, 0)

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

?‍♂️solution – (i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

Given –
The parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.
By using the formula,
The distance (d) between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is given by

∴ The distance between parallel lines is 65/17
(ii) l(x + y) + p = 0 and l (x + y) – r = 0

Given –
The parallel lines are l (x + y) + p = 0 and l (x + y) – r = 0.
lx + ly + p = 0 and lx + ly – r = 0
by using the formula,
The distance (d) between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is given by

∴ The distance between parallel lines is |p+r|/l√2

7. Find equation of the line parallel to the line 3x − 4y + 2 = 0 and passing through the point (–2, 3).

?‍♂️solution – Given –
The line is 3x – 4y + 2 = 0
So, y = 3x/4 + 2/4
= 3x/4 + ½
Which is of the form y = mx + c, where m is the slope of the given line.
The slope of the given line is 3/4
We know that parallel line have same slope.
∴ Slope of other line = m = 3/4
Equation of line having slope m and passing through (x₁, y₁) is given by
y – y1 = m (x – x₁)
∴ Equation of line having slope 3/4 and passing through (-2, 3) is
y – 3 = ¾ (x – (-2))
4y – 3 × 4 = 3x + 3 × 2
3x – 4y = 18
∴ The equation is 3x – 4y = 18

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

?‍♂️solution – Given –
The equation of line is x – 7y + 5 = 0
So, y = 1/7x + 5/7 [which is of the form y = mx + c, where m is the slope of the given line.]
Slope of the given line is 1/7
Slope of the line perpendicular to the line having slope m is -1/m
Slope of the line perpendicular to the line having a slope of 1/7 is -1/(1/7) = -7
So, the equation of 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
∴ The equation is 7x + y = 21

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

?‍♂️solution – Given –
The lines are √3x + y = 1 and x + √3y = 1
So, y = -√3x + 1 … (1) and
y = -1/√3x + 1/√3 …. (2)
Slope of line (1) is m1 = -√3, while the slope of line (2) is m₂ = -1/√3
Let θ be the angle between two lines
So,

θ = 30°
∴ The angle between the given lines is either 30° or 180°- 30° = 150°

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.

?‍♂️solution – Let the slope of the line passing through (h, 3) and (4, 1) be m1
Then, m1 = (1-3)/(4-h) = -2/(4-h)
Let the slope of line 7x – 9y – 19 = 0 be m₂
7x – 9y – 19 = 0
So, y = 7/9x – 19/9
m₂ = 7/9
Since, the given lines are perpendicular
m₁ × m₂= -1
-2/(4-h) × 7/9 = -1
-14/(36-9h) = -1
-14 = -1 × (36 – 9h)
36 – 9h = 14
9h = 36 – 14
h = 22/9
∴ The value of h is 22/9

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.

?‍♂️solution – Let the slope of line Ax + By + C = 0 be m
Ax + By + C = 0
So, y = -A/Bx – C/B
m = -A/B
By using the formula,
Equation of the line passing through point (x₁, y₁) and having slope m = -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
So, the line through point (x₁, y₁) and parallel to the line Ax + By + C = 0 is A (x – x₁) + B (y – y₁) = 0
Hence proved.

12. Two lines passing through the point (2, 3) intersects each other at an angle of 60^o. If slope of one line is 2, find equation of the other line.

?‍♂️solution – Given: m₁ = 2
Let the slope of the first line be m1
And let the slope of the other line be m2.
Angle between the two lines is 60°.
So,

13. Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).

?‍♂️solution – Given –
The right bisector of a line segment bisects the line segment at 90°.
End-points of the line segment AB are given as A (3, 4) and B (–1, 2).
Let mid-point of AB be (x, y)
x = (3-1)/2= 2/2 = 1
y = (4+2)/2 = 6/2 = 3
(x, y) = (1, 3)
Let the slope of line AB be m₁
m₁ = (2 – 4)/(-1 – 3)
= -2/(-4)
= 1/2
And let the slope of the line perpendicular to AB be m₂
m₂ = -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
∴ The required equation of the line is 2x + y = 5

14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

?‍♂️solution – Let us consider the co-ordinates of the foot of the perpendicular from (-1, 3) to the line 3x – 4y – 16 = 0 be (a, b)
So, let the slope of the line joining (-1, 3) and (a, b) be m₁
m₁ = (b-3)/(a+1)
And let the slope of the line 3x – 4y – 16 = 0 be m₂
y = 3/4x – 4
m₂ = 3/4
Since these two lines are perpendicular, m₁ × m₂ = -1
(b-3)/(a+1) × (3/4) = -1
(3b-9)/(4a+4) = -1
3b – 9 = -4a – 4
4a + 3b = 5 …….(1)
Point (a, b) lies on the line 3x – 4y = 16
3a – 4b = 16 ……..(2)
Solving equations (1) and (2), we get
a = 68/25 and b = -49/25
∴ The co-ordinates of the foot of perpendicular is (68/25, -49/25)

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.

?‍♂️solution – Given –
The perpendicular from the origin meets the given line at (–1, 2).
The equation of line is y = mx + c
The line joining the points (0, 0) and (–1, 2) is perpendicular to the given line.
So, the slope of the line joining (0, 0) and (–1, 2)=
2/(-1) = -2
Slope of the given line is m.
m × (-2) = -1
m = 1/2
Since, point (-1, 2) lies on the given line,
y = mx + c
2 = 1/2 × (-1) + c
c = 2 + 1/2 = 5/2
∴ The values of m and c are 1/2 and 5/2 respectively.

16. If p and q are the lengths of 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

?‍♂️solution – Given –
The equations of given lines are
x cos θ – y sin θ = k cos 2θ …………………… (1)
x sec θ + y cosec θ = k ……………….… (2)
Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

q = k cos θ sin θ
Multiply both sides by 2, we get
2q = 2k cos θ sin θ = k × 2sin θ cos θ
2q = k sin 2θ
Squaring both sides, we get
4q² = k² sin²2θ …………………(4)
Now add (3) and (4) we get
p² + 4q² = k² cos² 2θ + k² sin² 2θ
p² + 4q² = k² (cos² 2θ + sin² 2θ) [Since, cos² 2θ + sin² 2θ = 1]
∴ p² + 4q² = k²
Hence proved.

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.

?‍♂️solution – 

Let AD be the altitude of triangle ABC from vertex A.
So, AD is perpendicular to BC
Given:
Vertices A (2, 3), B (4, –1) and C (1, 2)
Let slope of line BC = m₁
m₁ = (- 1 – 2)/(4 – 1)
m₁ = -1
Let slope of line AD be m₂
AD is perpendicular to BC
m1 × m₂ = -1
-1 × m₂ = -1
m₂ = 1
The equation of the line passing through point (2, 3) and having a slope of 1 is
y – 3 = 1 × (x – 2)
y – 3 = x – 2
y – x = 1
Equation of the altitude from vertex A = y – x = 1
Length of AD = Length of the perpendicular from A(2, 3) to BC
Equation of BC is
y + 1 = -1 × (x – 4)
y + 1 = -x + 4
x + y – 3 = 0 …………………(1)
Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

Now compare equation (1) to the general equation of line i.e., Ax + By + C = 0, we get

Length of AD =

[where, A = 1, B = 1 and C = -3]
∴ The equation and the length of the altitude from
vertex A are y – x = 1 and
√2 units respectively.

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 1/p² = 1/a² + 1/b² 

?‍♂️solution – Equation of a line whose intercepts on the axes are a and b is x/a + y/b = 1
bx + ay = ab
bx + ay – ab = 0 ………………..(1)
Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

Now square on both the sides we get

Hence proved.