Question

Let A = (2, 3, 5), B = (– 1, 3, 2) and C = (a, 5, b) be the vertices of triangle ABC. If the median through A is equally inclined to the coordinate axes, then

• A. 5a – 8b = 0
• B. 8a – 5b = 0
• C. 10a – 7b = 0
• D. 7a – 10b = 0

Answer 1

The correct option is C 10a – 7b = 0

Mid point of $BC=(\frac{-1+9}{2},4,\frac{2+b}{2})=M$
A =(2, 3, 5)
dr's of AM =$(\frac{a-5}{2},1,\frac{b-8}{2})$=given AM is equally inclined to axes
$\Rightarrow \frac{a-5}{2}=1=\frac{b-8}{2}\Rightarrow a=7b=10\Rightarrow 10a-7b=0$