Question

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

• A. the equation of line $BC$ is $7x-3y+4=0$
• B. the equation of line $BC$ is $7x+3y+4=0$
• C. $B\equiv (\frac{5}{2},\frac{9}{2})$
• D. $B\equiv (-\frac{5}{2},\frac{9}{2})$

Answer 1

Answer: (B), (D)

The correct options are
B the equation of line $BC$ is $7x+3y+4=0$
D $B\equiv (-\frac{5}{2},\frac{9}{2})$


Let the image of $A(1,3)$ w.r.t. $x+y=2$ be $D(a,b)$, so
$\frac{a-1}{1}=\frac{b-3}{1}=-\frac{2(1+3-2)}{2}\Rightarrow D=(-1,1)$

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

Therefore, vertex $B$ is point of the intersection of $7x+3y+4=0$ and $x+y=2$
i.e. $B\equiv (-\frac{5}{2},\frac{9}{2})$