Question

Find the equation of a line whose gradient is `-1/3` and which passes through a point M which divides the line segment joining A(8, −4) and B(0, −12) in the ratio 1 : 3.

Answer 1

The coordinates of a point dividing a line segment A(x₁, y₁) and B(x₂, y₂) in the ratio m : n are:

`M(x, y) = ((mx_2 + nx_1)/(m + n), (my_2 + ny_1)/(m + n))`

Given:

• A = (x₁, y₁) = (8, −4)
• B = (x₂, y₂) = (0, −12)
• Ratio m : n = 1 : 3

Calculation of x:

⇒ `x = (1(0) + 3(8))/(1 + 3)`

`x = 24/4`

∴ x = 6

⇒ `y = (1(-12) + 3(-4))/(1 + 3)`

`y = (-12 - 12)/4`

`y = (-24)/4`

∴ y = −6

So, the coordinates of point M are (6, −6).

Given:

• Gradient (m) = `-1/3`
• Passing through point M(6, −6)

Using the point-slope formula:

y − y₁ = m(x − x₁)

`y - (-6) = -1/3(x - 6)`

`y + 6 = -1/3(x - 6)`

Multiply the entire equation by 3 to clear the fraction:

3(y + 6) = −1(x − 6)

3y + 18 = −x + 6

Rearrange in standard form (Ax + By + C = 0):

x + 3y + 18 − 6 = 0

x + 3y + 12 = 0

Hence, the equation of the line is x + 3y + 12 = 0.