Chapter 10: Straight Lines Exercise 10.3

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.

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,

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

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)

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.

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.

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 =

∴ The equation and the length of the altitude from vertex A are y – x = 1 and

√2 units respectively.

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

∴ 1/p² = 1/a² + 1/b²

Hence proved.