Question

P is a point on the x-axis which divides the line joining A(–6, 2) and B(9, –4). Find:

1. the ratio in which P divides the line segment AB.
2. the coordinates of the point P.
3. equation of a line parallel to AB and passing through (–3, –2).

Answer 1

a. Let the point P(x, 0) divide the line segment joining A(–6, 2) and B(9, –4) in the ratio m : n.

By section formula,

`y = (my_2 + ny_1)/(m + n)`

Substituting the values, we get:

⇒ `0 = (m(-4) + n(2))/(m + n)`

⇒ 0 = – 4m + 2n

⇒ 4m = 2n

⇒ `m/n = 2/4`

⇒ `m/n = 1/2`

Hence, the ratio in which P divides the line segment AB is 1 : 2.

b. From part (a),

A(–6, 2) and B(9, –4)

m : n = 1 : 2

By section-formula,

`x = (mx_2 + nx_1)/(m + n)`

= `(1(9) + 2(-6))/(1 + 2)`

= `(9 - 12)/3`

= `(-3)/3`

= –1

P = (x, 0) = (–1, 0)

Hence, the coordinates of the point P are (–1, 0).

c. By formula,

`m = (y_2 - y_1)/(x_2 - x_1)`

Substituting values, we get:

⇒ `m_(AB) = (-4 - 2)/(9 - (-6))`

⇒ `m_(AB) = (-6)/15`

⇒ `m_(AB) = (-2)/5`

Line parallel to AB will have the same slope, so the slope of the new line is also `m = (-2)/5`.

By point-slope formula,

y – y₁ = m(x – x₁)

Line passing through (–3, –2) and parallel to AB is:

⇒ `y - (-2) = (-2)/5(x - (-3))`

⇒ `y + 2 = (-2)/5(x + 3)`

⇒ 5(y + 2) = –2(x + 3)

⇒ 5y + 10 = –2x – 6

⇒ 5y + 10 + 2x + 6 = 0

⇒ 2x + 5y + 16 = 0

Hence, the equation of the line is 2x + 5y + 16 = 0.