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.

Let the co-ordinates of P be (x, y)

$\therefore x = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] = \dfrac{3 \times 5 + 2 \times 0}{3 + 2} \\[1em] = \dfrac{15 + 0}{3} \\[1em] = \dfrac{15}{3} = 5.

and,

$y = \dfrac{m$1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{3 \times 10 + 2 \times 5}{3 + 2} \\[1em] = \dfrac{30 + 10}{5} \\[1em] = \dfrac{40}{5} = 8.

P = (x, y) = (5, 8).

Distance between two points = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

$AP = \sqrt{(5 - 4)^2 + (8 - (-4))^2} \\[1em] = \sqrt{1^2 + 12^2} \\[1em] = \sqrt{1 + 144} \\[1em] = \sqrt{145} \\[1em] = 12.04$

Hence, AP = 12.04 units.