Question

Determine the ratio in which the line 3x + y – 9 = 0 divides the line segment joining the points (1, 3) and (2, 5). Find the point of intersection.

Answer 1

Let A(1,3), B(2,5).

Suppose the line meets AB at P, dividing it internally in the ratio AP: PB = k:1.

By the section formula, P = `((k·x2 + x1)/(k + 1), (k·y2 + y1)/(k + 1)).`

Coordinates of P are x = `(2k + 1)/(k + 1), y = (5k + 3)/(k + 1)`

Substitute into 3x + y − 9 = 0 and multiply by (k + 1)

 `3(2k + 1) + (5k + 3) − 9(k + 1) = 0`

⇒ 6k + 3 + 5k + 3 − 9k − 9 = 0

⇒ 2k − 3 = 0

⇒ k = `3/2`.

So AP: PB = 3:2.

Find P using k = `3/2`

x = `(2·(3/2) + 1)/(3/2 + 1)`

= `4/(5/2)`

= `8/5, y`

= `(5·(3/2) + 3)/(3/2 + 1)`

= `(21/2)/(5/2)`

= `21/5`

The line divides the segment in the ratio 3:2 (internally).

The point of intersection is P = `(8/5, 21/5)`