Question

The equation of the internal bisector of $\angle BAC$ of the triangle ABC whose vertices are $A(5,2),B(2,3),C(6,5)$ is

• A. $6x+y-32=0$
• B. $x-4y+3=0$
• C. $2x+y-12=0$
• D. $y-2=0$

Answer 1

Answer: (C)

$2x+y-12=0$

Given vertices

$A(5,2),B(2,3),C(6,5)$

Eq of line $AB$ from Points A and B

$y-2=\frac{3-2}{2-5}(x-5)$

$y-2=\frac{1}{-3}(x-5)$

$-3y+6=x-5$

$x+3y-11=0$-----(1)

Eq of line $AC$ from Points A and C

$y-2=\frac{5-2}{6-5}(x-5)$

$y-2=\frac{3}{1}(x-5)$

$y-2=3x-15$

$3x-y-13=0$-----(2)

Eq of angle bisector of AB and AC

$x+3y-11=\pm (3x-y-13)$

$x+3y-11=3x-y-13$ and $x+3y-11=-3x+y+13$

$2x-4y-2=0$ and $4x+2y-24=0$

$x-2y-1=0$ and $2x+y-12=0$