Question

If $A(2,2,-3),B(5,6,9),C(2,7,9)$ be the vertices of a triangle. The internal bisector of the angle $A$ meets $BC$ at the point $D$, then find the coordinates of $D$.

• A. $(\frac{13}{2},\frac{7}{2},9)$
• B. $(\frac{7}{2},\frac{13}{2},9)$
• C. $(\frac{9}{2},\frac{7}{2},9)$
• D. $(\frac{13}{2},\frac{9}{2},9)$

Answer 1

The correct option is B $(\frac{7}{2},\frac{13}{2},9)$

We have,

Point

$A(x_1,y_1,z_1)=(2,2,-3)$

$B(x_2,y_2,z_2)=(5,6,9)$

$C(x_3,y_3,z_3)=(2,7,9)$

Let the coordinates of point $D(x,y,z).$

So,

According to question,

$AB=\sqrt{{(5-2)}^2+{(6-2)}^2+{(9-3)}^2}$

$AB=\sqrt{9+16+149}=\sqrt{{13}^2}=13$

$AC=\sqrt{{(2-2)}^2+{(7-2)}^2+{(9+3)}^2}$

$AC=\sqrt{0+25+144}=\sqrt{169}=13$

Thus $ABC$ is isosceles triangle with $AB=AC$

So, angle bisector $AD$ bisects $BC$

$D\equiv (\frac{5+2}{2},\frac{6+7}{2},\frac{9+9}{2})\equiv (\frac{7}{2},\frac{13}{2},9)$

Hence, this is the ansewr.