Question

Let $ABC$ be triangle. Let $A$ be the point $(1,2),y=x$be the perpendicular bisector of $AB$ and $x-2y+1=0$ be the angle bisector of $\angle C$. If equation of $BC$ is given by $ax+by-5=0$, then the value of $a+b$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$\begin{array}{c}{m}_{AB}=-1\\ y-2=-1\left(x-1\right)\\ y-2=-x+1\\ x+y=3\\ \underset{–}{x-y=0}\\ 2x=3\\ \therefore x=\frac{3}{2},y=\frac{3}{2}\\ \frac{1+h}{2}=\frac{3}{2}\\ h=2\\ \frac{2+k}{2}=\frac{3}{2}\\ k=1B\left(2,1\right)\\ Asimagethroughx-2y+1=0\\ m_{AD}=-2\\ y-2=-2\left(x-1\right)\\ y-2=-2x+2\\ 2x+y=4\times 2x-2y+1=0\\ 4x+2y=8\\ \underset{–}{x-2y=-1}\\ 5x=7\\ x=\frac{7}{5}y=\frac{x+1}{2}=\frac{12}{5\times 2}=\frac{6}{5}\\ \frac{m+1}{2}=\frac{7}{5}\frac{n+1}{2}=\frac{6}{5}\\ m+1=\frac{14}{5}n+2=\frac{12}{5}\\ m=\frac{9}{5}n=\frac{12}{5}-2=\frac{2}{5}\\ E\left(\frac{9}{5},\frac{2}{5}\right)\\ BC=BEB\left(2,1\right)E\left(\frac{9}{5},\frac{2}{5}\right)\\ y-1=\left(\frac{1-\frac{2}{5}}{2-\frac{9}{5}}\right)\left(x-2\right)\\ 3x-y=5\\ a=3b=-1\\ a+b=2\end{array}$