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.
"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)."
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