Question

If A is the point (−3, 5, 10) and B is the point (12, −5, −15), find the coordinates of point P on the line AB given that AP: PB = 2: 3.

Answer 1

$$x_P=\dfrac{2x_B+3x_A}{2+3}=\dfrac{2(12)+3(-3)}{5}=3$$

$$y_P=\dfrac{2y_B+3y_A}{2+3}=\dfrac{2(-5)+3(5)}{5}=1$$

$$z_P=\dfrac{2z_B+3z_A}{2+3}=\dfrac{2(-15)+3(10)}{5}=0$$

P is the point $(3,1,0).$