Question

$A(1,3)$ and $C(-2/5,-2/5)$ are the vertices of a triangle $ABC$ and the equation of the internal angle bisector of $\angle ABC$ is $x+y=2$, then

• A. $BC:7x-3y+4=0$
• B. $BC:7x+3y+4=0$
• C. $B\equiv (5/2,9/2)$
• D. $B\equiv (-5/2,9/2)$

Answer 1

Answer: (B), (D)

The correct options are
B $BC:7x+3y+4=0$
D $B\equiv (-5/2,9/2)$


Let the image of $A(1,3)$ in line $x+y=2$ be $D(a,b)$. Here the midpoint of $AD$ which is $(\frac{a+1}{2},\frac{b+3}{2})$ lies on the line $x+y=2$.

Substituting this point in the line equation $x+y=2$, we get $(a,b)=(-1,1)$

So line $BC$ passes through $(-1,1)$ and $(-2/5,-2/5).$

The equation of line $BC$ is
$y-1=\frac{-2/5-1}{-2/5+1}(x+1)$
$\Rightarrow 7x+3y+4=0$

Vertex $B$ is point of intersection of $7x+3y+4=0$ and $x+y=2,$ i.e., $B\equiv (-5/2,9/2)$