Question

$A(3,2,0),B(5,3,2)$ and $C(-9,6,-3)$ are the vertices of triangle $ABC$. If the bisector of $\angle ABC$ meets $BC$ at $D$, then coordinates of $D$ are:

• A. $(19/8,57/16,17/16)$
• B. $(-19/8,57/16,17/16)$
• C. $(19/8,-57/16,17/16)$
• D. $Noneofthese$

Answer 1

Answer: (A)

$(19/8,57/16,17/16)$

Solution:- (A) $(\frac{19}{8},\frac{57}{16},\frac{17}{16})$

Given that:- $ABC$ is a triangle and bisector $AD$ of $\angle A$ meets $BC$ at $D$.

To find:- Coordinates of point $D$

Solution:-

Since $AD$ is the bisector of $\angle BAC$

$\therefore \frac{BD}{DC}=\frac{AB}{AC}.....\left(1\right)$

As the vertices of $A,B$ and $C$ of $△ABC$ are given.

Therefore,

$AB=\sqrt{{(5-3)}^2+{(3-2)}^2+{(2-0)}^2}=3$

$AC=\sqrt{{((-9)-3)}^2+{(6-2)}^2+{((-3)-0)}^2}=13$

Substituting the values of $AB$ and $AC$ in ${eq}^n\left(1\right)$, we have

$\frac{BD}{DC}=\frac{3}{13}$, i.e.,

$d$ divides $BC$ in ration $3:13$.

Therefore, coordinates of $D$ are-

$(\frac{3(-9)+13\left(5\right)}{3+13},\frac{3\left(6\right)+13\left(3\right)}{3+13},\frac{3(-3)+13\left(2\right)}{3+13})$

$\Rightarrow (\frac{19}{8},\frac{57}{16},\frac{17}{16})$