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 $BC$ 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 $AD$ is the internal bisector of $\angle A$. Then, we have (by geometry)
$\frac{BD}{CD}=\frac{AB}{AC}$

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