Question

The vertex A of a triangle ABC is the point $(-2,3)$ whereas the line perpendicular to the sides AB and AC are $x-y-4=0$ and $2x-y-5=0$ respectively. The right bisectors of sides meet at P$(3/2,5/2)$ . Then the equation of side BC is

• A. $5x+2y=10$
• B. $5x-2y=16$
• C. $2x-5y=10$
• D. none of these

Answer 1

The correct option is D none of these

Given vertex of $\Delta ABC$

$A(-2,3)$

Equation of side AB perpendiuclar to $x-y-4=0$ is

$AB:x+y-\lambda =0$

Side AB passes through point A$(-2,3)$

$-2+3-\lambda =0$

$\lambda =1$

Hence $AB:x+y-1=0----\left(1\right)$

Equation of side AC perpendiuclar to $2x-y-5=0$ is

$AC:x+2y-\lambda =0$

Side AC passes through point A$(-2,3)$

$-2+6-\lambda =0$

$\lambda =4$

Hence $AC:x+2y-4=0----\left(2\right)$

The right bisectors of all sides are meeting at point $P(\frac{3}{2},\frac{5}{2})$

Right bisector from point B on side AC is perpendicular

Hence Equation of right bisector perpendicular to $x+2y-4=0$ is

$2x-y-\lambda =0$

Here right bisector is passing through point P

$2\times \frac{3}{2}-\frac{5}{2}\lambda =0$

$\frac{6}{2}-\frac{5}{2}=\lambda$

$\lambda =\frac{1}{2}$

Equation of right bisector $BD$ be

$2x-y-\frac{1}{2}=0$

$4x-2y-1=0----\left(3\right)$

Here On solving equation of $AB$ and $BD$ we get point of intersection i.e. point $B$

$4(1-y)-2y-1=0$

$4-4y-2y-1=0$

$-6y=-3$

$y=\frac{1}{2}$

$x=1-y=1-\frac{1}{2}=\frac{1}{2}$

Point $B(\frac{1}{2},\frac{1}{2})$

Right bisector from point A on side BC is passing through point P

Hence equation of AD from point A and P

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

$y-3=\frac{-1}{7}(x+2)$

Slope of AD is $m_{AD}=\frac{-1}{7}$

Hence slope of side $BC$ perpendicular to AD is $-\frac{1}{7}{m}_{BC}=-1$

$m_{BC}=7$

Equation of side BC from point B$(\frac{1}{2},\frac{1}{2})$ with slope $m_{BC}=7$

$y-\frac{1}{2}=7(x-\frac{1}{2})$

$2y-1=14x-7$

$14x-2y-6=0$