Question

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

• A. $(\frac{19}{8},\frac{57}{16},\frac{17}{16})$
• B. $(-\frac{19}{8},\frac{57}{16},\frac{17}{16})$
• C. $(\frac{19}{8},\frac{57}{16},-\frac{17}{16})$
• D. $(-\frac{19}{8},\frac{57}{16},-\frac{17}{16})$

Answer 1

The correct option is A $(\frac{19}{8},\frac{57}{16},\frac{17}{16})$

R.E.F image

In $\Delta ABC,$

AD is the bisector of

$\angle A\Delta$ meets BC at D.

$\therefore \frac{BD}{DC}=\frac{AB}{AC}$

or

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

$=\sqrt{2^2+1^2+2^2}=\sqrt{9}$

$=3$

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

$=\sqrt{{12}^2+4^2+3^2}$

$=\sqrt{144+16+9}$

$=13$

$\frac{BD}{DC}=\frac{3}{13}$

Coordinate of D $\Rightarrow$

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

$=(\frac{19}{8},\frac{57}{16},\frac{17}{16})$ option A