Chapter 10: Straight Lines Exercise 10.2

The y-coordinate of every point on x-axis is 0.

∴ Equation of x-axis is y = 0.

The x-coordinate of every point on y-axis is 0.

∴ Equation of y-axis is y = 0.

Given:

Point (-4, 3) and slope, m = 1/2

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

So, y – 3 = 1/2 (x – (-4))

y – 3 = 1/2 (x + 4)

2(y – 3) = x + 4

2y – 6 = x + 4

x + 4 – (2y – 6) = 0

x + 4 – 2y + 6 = 0

x – 2y + 10 = 0

∴ The equation of the line is x – 2y + 10 = 0.

Given:

Point (0, 0) and slope, m = m

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

So, y – 0 = m (x – 0)

y = mx

y – mx = 0

∴ The equation of the line is y – mx = 0.

Given: point (2, 2√3) and θ = 75°

Equation of line: (y – y₁) = m (x – x₁)

where, m = slope of line = tan θ and (x₁, y₁) are the points through which line passes

∴ m = tan 75°

75° = 45° + 30°

Applying the formula:

We know that the point (x, y) lies on the line with slope m through the fixed point (x₁, y1), if and only if, its coordinates satisfy the equation y – y₁ = m (x – x₁)

Then, y – 2√3 = (2 + √3) (x – 2)

y – 2√3 = 2 x – 4 + √3 x – 2 √3

y = 2 x – 4 + √3 x

(2 + √3) x – y – 4 = 0

∴ The equation of the line is (2 + √3) x – y – 4 = 0.

Given:

Slope, m = -2

We know that if a line L with slope m makes x-intercept d, then equation of L is

y = m(x − d).

If the distance is 3 units to the left of origin then d = -3

So, y = (-2) (x – (-3))

y = (-2) (x + 3)

y = -2x – 6

2x + y + 6 = 0

∴ The equation of the line is 2x + y + 6 = 0.

Given: θ = 30°

We know that slope, m = tan θ

m = tan30° = (1/√3)

We know that the point (x, y) on the line with slope m and y-intercept c lies on the line if and only if y = mx + c.

If distance is 2 units above the origin, c = +2

So, y = (1/√3)x + 2

y = (x + 2√3) / √3

√3 y = x + 2√3

x – √3 y + 2√3 = 0

∴ The equation of the line is x – √3 y + 2√3 = 0.

Given:

Points (-1, 1) and (2, -4)

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

y – 1 = -5/3 (x + 1)

3 (y – 1) = (-5) (x + 1)

3y – 3 = -5x – 5

3y – 3 + 5x + 5 = 0

5x + 3y + 2 = 0

∴ The equation of the line is 5x + 3y + 2 = 0.

Given: p = 5 and ω = 30°

We know that the equation of the line having normal distance p from the origin and angle ω which the normal makes with the positive direction of x-axis is given by x cos ω + y sin ω = p.

Substituting the values in the equation, we get

x cos30° + y sin30° = 5

x(√3 / 2) + y( 1/2 ) = 5

√3 x + y = 5(2) = 10

√3 x + y – 10 = 0

∴ The equation of the line is √3 x + y – 10 = 0.

Given:

Vertices of ΔPQR i.e. P (2, 1), Q (-2, 3) and R (4, 5)

Let RL be the median of vertex R.

So, L is a midpoint of PQ.

We know that the midpoint formula is given by
.

∴ L = = (0, 2)

We know that the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is given by

y – 5 = -3/-4 (x-4)

(-4) (y – 5) = (-3) (x – 4)

-4y + 20 = -3x + 12

-4y + 20 + 3x – 12 = 0

3x – 4y + 8 = 0

∴ The equation of median through the vertex R is 3x – 4y + 8 = 0.

Given:

Points are (2, 5) and (-3, 6).

We know that slope, m = (y₂ – y₁)/(x₂ – x₁)

= (6 – 5)/(-3 – 2)

= 1/-5 = -1/5

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Then, m = (-1/m)

= -1/(-1/5)

= 5

We know that the point (x, y) lies on the line with slope m through the fixed point (x₀, y₀), if and only if, its coordinates satisfy the equation y – y₀ = m (x – x₀)

Then, y – 5 = 5(x – (-3))

y – 5 = 5x + 15

5x + 15 – y + 5 = 0

5x – y + 20 = 0

∴ The equation of the line is 5x – y + 20 = 0