Question

If the vertices of $△ABC$ are $A=(3,-5),B=(-3,3)$ and $C=(-1,-2)$ and internal angle bisector through $A$ cuts the side $BC$ at $D,$ then the length of $AD$ is

• A. $\frac{7}{3}\sqrt{2}$ units
• B. $\frac{14}{3}\sqrt{2}$ units
• C. $\frac{3}{7}\sqrt{2}$ units
• D. $\frac{3}{14}\sqrt{2}$ units

Answer 1

The correct option is B $\frac{14}{3}\sqrt{2}$ units

In $\Delta ABC$, bisector through $A$ divides opposide side in the ratio of other two sides.
So, here $\frac{BD}{DC}=\frac{AB}{AC}=\frac{c}{b}=\frac{10}{5}=2$

Now, $D$ divides $BC$ in the ratio $2:1$
So, $D=(\frac{2(-1)+1(-3)}{2+1},\frac{2(-2)+1\left(3\right)}{2+1})$
$=(\frac{-5}{3},\frac{-1}{3})$

Now, $AD=\sqrt{{\left(\frac{14}{3}\right)}^2+{\left(\frac{-14}{3}\right)}^2}$
$\therefore AD=\frac{14}{3}\sqrt{2}$ units