Question

Given a triangle ABC in which A = (4, −4), B = (0, 5) and C = (5, 10). A point P lies on BC such that BP : PC = 3 : 2. Find the length of line segment AP.

Answer 1

Given, BP : PC = 3 : 2

Using section formula, the co-ordinates of point P are

`((3 xx 5 + 2 xx 0)/(3 + 2), (3 xx 10 + 2 xx 5)/(3 + 2))`

= `(15/5, 40/5)`

= (3, 8)

Using distance formula, we have:

`AP = sqrt((3 - 4)^2 + (8 + 4)^2)`

= `sqrt(1 + 144)`

= `sqrt(145)`

= 12.04