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

Using the section formula, if a point (x,y,z) divides the line joining the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_3)$ in the ratio $m:n$, then;

$(x,y,z) = (\dfrac{mx_2+ nx_1}{m+n}, \dfrac{my_2+ ny_1}{m+n}, \dfrac{mz_2+ nz_1}{m+n})\\ \quad\\ AP: PB = 2: 3\\ m: n = 2:3\\ \quad\\ (x,y,z) = (\dfrac{2(12)+ 3(-3)}{2+3}, \dfrac{2(-5)+ 3(5)}{3+2}, \dfrac{2(-15)+ 3(10)}{3+2})\\ \quad\\ (x,y,z) = (\dfrac{24-9}{5}, \dfrac{-10+15}5, \dfrac{-30+30} 5)\\ (x,y,z) = ( 3,1,0)\\ \quad\\ P \implies (x,y) = (3,1)$