Question

Let $ABC$ be a 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 $2$

The slope of the line AB will be $-1$
Therefore the equation of the line AB will be $\frac{y-2}{x-1}=-1$
$y-2=-x+1$
$x+y=3$
Since $y=x$ is the perpendicular bisector of AB, therefore, the midpoint of AB is given by the solution set of $y=x$ and the equation of AB
Hence by solving the above pair of linear equations, we get the mid point as $\frac{3}{2},\frac{3}{2}$
Hence the coordinates of B is $(2,1)$
Taking the image of point A on the angle bisector of C we get
$\frac{x-1}{1}=\frac{y-2}{-2}+\frac{4}{5}$
Therefore the coordinates of the point will be $(\frac{9}{5},\frac{2}{5})$
Hence the equation of BC will be
$3x-y=5$
Therefore $a+b=2$