Question

If $A(2,2,-3), B(5,6,9)$ and $C(2,7,9)$ are the vertices of a triangle and the internal bisector of the $\angle A$ meets $B C$ at the point $D$, then the coordinates of $D$ are

• A. $\left(\frac{3}{2}, \frac{13}{2}, 9\right)$
• B. $\left(\frac{7}{2}, \frac{13}{2}, 9\right)$
• C. $\left(\frac{5}{2}, \frac{7}{2}, 9\right)$
• D. None of these

Answer 1

Answer: (B)

If $A D$ is the internal bisector of $\angle A$. Then, we have (by geometry)
$\frac{B D}{C D}=\frac{A B}{A C}$

$\frac{A B}{A C} =\frac{\sqrt{(2-5)^2+(2-6)^2+(-3-9)^2}}{\sqrt{(2-2)^2+(2-7)^2+(-3-9)^2}}$
$ =\frac{\sqrt{9+16+144}}{\sqrt{0+25+144}}=\frac{1}{1}$
Therefore, $D$ is the mid-point of $B C$.
$\therefore D=\left(\frac{5+2}{2}, \frac{6+7}{2}, \frac{9+9}{2}\right)=\left(\frac{7}{2}, \frac{13}{2}, 9\right)$