Question

$A(3,2,0),B(5,3,2),C(-9,6,-3)$ are three points forming a triangle. If AD, the bisector of $\angle BAC$ meets BC in D, then coordinates of D 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 D

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

We known that

The bisector of an angle of a triangle divides the opposite

side into segments that are proportional to the adjacent sides.

That is, for a triangle ABC,the bisector of the angle at A divides

the opposite side into segment of length BD and DC.

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

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

$=\sqrt{4+1+4}=\sqrt{9}=3$

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

$=\sqrt{144+16+9}=\sqrt{169}=13$

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

$\Rightarrow$ D divides BC in the ratio 3:13

Coordinates of the point D

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

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

Hence,option D is correct answer