Question

A (–3, 4), B (3, –1) and C (–2, 4) are the vertices of a triangle ABC. Find the length of line segment AP, where point P lies inside BC, such that BP : PC = 2 : 3.

Answer 1

BP : PC = 2 : 3

Co-ordinates of P are

`((2 xx (-2) + 3 xx 3)/(2 + 3),(2 xx 4 + 3 xx (-1))/(2 + 3))`

= `((-4 + 9)/5, (8 - 3)/5)`

= (1, 1)  ...(i)

Using distance formula, we have:

`AP = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`

Let A (–3 , 4)  x₁ = –3, y₁ = 4,

(1, 1) x₂ = 1, y₂ = 1   ...[From (i) we get]

`AP = sqrt((1 + 3)^2 + (1 - 4)^2)`  

= `sqrt(16 + 9)`

= `sqrt(25)`

= 5 units.