Chapter 10: Straight Lines Exercise 10.2

We know that the coordinates of a point dividing the line segment joining the points (x₁, y₁) and (x₂, y₂) internally in the ratio m: n are

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

= (3 – 0)/(2 – 1)

= 3/1

= 3

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) = -1/3

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)

Here, the point is

3((1 + n) y – 3) = (-(1 + n) x + 2 + n)

3(1 + n) y – 9 = – (1 + n) x + 2 + n

(1 + n) x + 3(1 + n) y – n – 9 – 2 = 0

(1 + n) x + 3(1 + n) y – n – 11 = 0

∴ The equation of the line is (1 + n) x + 3(1 + n) y – n – 11 = 0.

Given: the line cuts off equal intercepts on the coordinate axes i.e. a = b.

We know that equation of the line intercepts a and b on x-and y-axis, respectively, which is

x/a + y/b = 1

So, x/a + y/a = 1

x + y = a … (1)

Given: point (2, 3)

2 + 3 = a

a = 5

Substitute value of ‘a’ in (1), we get

x + y = 5

x + y – 5 = 0

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

We know that equation of the line making intercepts a and b on x-and y-axis, respectively, is x/a + y/b = 1 . … (1)

Given: sum of intercepts = 9

a + b = 9

b = 9 – a

Now, substitute value of b in the above equation, we get

x/a + y/(9 – a) = 1

Given: the line passes through the point (2, 2),

So, 2/a + 2/(9 – a) = 1

18/a(9 – a) = 1

18 = a (9 – a)

18 = 9a – a²

a² – 9a + 18 = 0

Upon factorizing, we get

a² – 3a – 6a + 18 = 0

a (a – 3) – 6 (a – 3) = 0

(a – 3) (a – 6) = 0

a = 3 or a = 6

Let us substitute in (1),

Case 1 (a = 3):

Then b = 9 – 3 = 6

x/3 + y/6 = 1

2x + y = 6

2x + y – 6 = 0

Case 2 (a = 6):

Then b = 9 – 6 = 3

x/6 + y/3 = 1

x + 2y = 6

x + 2y – 6 = 0

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

Given:

Point (0, 2) and θ = 2π/3

We know that m = tan θ

m = tan (2π/3) = -√3

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)

y – 2 = -√3 (x – 0)

y – 2 = -√3 x

√3 x + y – 2 = 0

Given, equation of line parallel to above obtained equation crosses the y-axis at a distance of 2 units below the origin.

So, the point = (0, -2) and m = -√3

From point slope form equation,

y – (-2) = -√3 (x – 0)

y + 2 = -√3 x

√3 x + y + 2 = 0

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

Given:

Points are origin (0, 0) and (-2, 9).

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

= (9 – 0)/(-2-0)

= -9/2

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

m = (-1/m) = -1/(-9/2) = 2/9

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)

y – 9 = (2/9) (x – (-2))

9(y – 9) = 2(x + 2)

9y – 81 = 2x + 4

2x + 4 – 9y + 81 = 0

2x – 9y + 85 = 0

∴ The equation of line is 2x – 9y + 85 = 0.

Let us assume ‘L’ along X-axis and ‘C’ along Y-axis, we have two points (124.942, 20) and (125.134, 110) in XY-plane.

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

Assuming the relationship between selling price and demand is linear.

Let us assume selling price per litre along X-axis and demand along Y-axis, we have two points (14, 980) and (16, 1220) in XY-plane.

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

y – 980 = 120 (x – 14)

y = 120 (x – 14) + 980

When x = Rs 17/litre,

y = 120 (17 – 14) + 980

y = 120(3) + 980

y = 360 + 980 = 1340

∴ The owner can sell 1340 litres weekly at Rs. 17/litre.

Let AB be a line segment whose midpoint is P (a, b).

Let the coordinates of A and B be (0, y) and (x, 0) respectively.

a (y – 2b) = -bx

ay – 2ab = -bx

bx + ay = 2ab

Divide both the sides with ab, then

Hence proved.

Let us consider, AB be the line segment such that r (h, k) divides it in the ratio 1: 2.

So the coordinates of A and B be (0, y) and (x, 0) respectively.

We know that the coordinates of a point dividing the line segment joins the points (x₁, y₁) and (x₂, y₂) internally in the ratio m: n is

h = 2x/3 and k = y/3

x = 3h/2 and y = 3k

∴ A = (0, 3k) and B = (3h/2, 0)

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

3h(y – 3k) = -6kx

3hy – 9hk = -6kx

6kx + 3hy = 9hk

Let us divide both the sides by 9hk, we get,

2x/3h + y/3k = 1

∴ The equation of the line is given by 2x/3h + y/3k = 1

According to the question,

If we have to prove that the given three points (3, 0), (– 2, – 2) and (8, 2) are collinear, then we have to also prove that the line passing through the points (3, 0) and (– 2, – 2) also passes through the point (8, 2).

By using the formula,

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

-5y = -2 (x – 3)

-5y = -2x + 6

2x – 5y = 6

If 2x – 5y = 6 passes through (8, 2),

2x – 5y = 2(8) – 5(2)

= 16 – 10

= 6

= RHS

The line passing through the points (3, 0) and (– 2, – 2) also passes through the point (8, 2).

Hence proved. The given three points are collinear.