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.

Given,

BP : PC = 2 : 3.

So, P divides the line segment BC in ratio 2 : 3.

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

$\therefore x = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] = \dfrac{2 \times -2 + 3 \times 3}{2 + 3} \\[1em] = \dfrac{-4 + 9}{5} \\[1em] = \dfrac{5}{5} = 1.

and

$y = \dfrac{m$1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{2 \times 4 + 3 \times -1}{2 + 3} \\[1em] = \dfrac{8 - 3}{5} \\[1em] = \dfrac{5}{5} = 1.

P = (x, y) = (1, 1).

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

$AP = \sqrt{[1 - (-3)]^2 + (1 - 4)^2} \\[1em] = \sqrt{(4)^2 + (-3)^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.$

Hence, AP = 5 units.