Question

The equations of the perpendicular bisector of the sides $AB$ and $AC$ of a $△ABC$ are $x-y+5=0$ and $x+2y=0$ respectively, if the point $A$ is $(1,-2)$, then the equation of the line $BC$ is

• A. $14x+2y-41=0$
• B. $11x+2y-25=0$
• C. $2x-y-10=0$
• D. None of these

Answer 1

The correct option is D None of these

Equation of the perpendicular bisector of $AB$ is $x-y+5=0$.........$\left(1\right)$

Also, Equation of $AB$ (is perpendicular to $x-y+5=0$) can be written as $y+x+\lambda =0$, but passes through $A=(1,-2)$

$\Rightarrow -2+1+\lambda =0$

$\Rightarrow \lambda =1$

$\therefore AB=y+x+1=0$ ........$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get

The coordinates of $D$, the middle point of $AB$ is

On solving $x-y=-5$

$x+y=-1$

$\Rightarrow x-y+x+y=-5-1$

$\Rightarrow 2x=-6$ or $x=-3$

Put $x=-3$ in $x+y=-1$ we get

$y=-1-x=-1-(-3)=-1+3=2$

$\therefore x=-3,y=2$

Hence the midpoint of $AB$ is $(-3,2)$

Now, $D$ is the middle point of $AB$,

Let the coordinates of $B$ be $(\alpha ,\beta )$

Then,

$\frac{\alpha +1}{2}=-3$ and $\frac{\beta -2}{2}=2$

$\Rightarrow \alpha +1=-6,\beta -2=4$

$\Rightarrow \alpha =-6-1=-7,\beta =4+2=6$

$\therefore$ co-ordinates of $B$ is $(-7,6)$ .........$\left(3\right)$

Now, equation of the line $AC,$

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

Solve this equation by $x+2y=0$ we get the coordinates of $E$ which is the middle point of $AC$

Substituting $x=-2y$ in $\left(4\right)$ we get

$\Rightarrow -4y-y-4=0$

$\Rightarrow -5y=4$ or $y=\frac{-4}{5}$

substitute $y=\frac{-4}{5}$ in $x=-2y$ we get

$x=-2\times \frac{-4}{5}=\frac{8}{5}$

Thus, the middle point of $AC$ is $E(\frac{8}{5},\frac{-4}{5})$

Let the coordinate of $C$ be $(p,q)$ , then

$\frac{p+1}{2}=\frac{8}{5}$ and $\frac{q+2}{2}=\frac{-4}{5}$

$\Rightarrow 5p+5=16$ and $5q+10=-8$

$\Rightarrow 5p=11$ and $5q=-18$

$\Rightarrow p=\frac{11}{5},q=\frac{-18}{5}$

$\Rightarrow C=(p,q)=(\frac{11}{5},\frac{-18}{5})$

Thus, the equation of $BC$ , which is passing through $B$ and $C$ , is given by

$y+\frac{18}{5}=\frac{6+\frac{18}{5}}{-7-\frac{11}{5}}(x-\frac{11}{5})$

$\Rightarrow y+\frac{18}{5}=\frac{30+18}{-35-11}(x-\frac{11}{5})$

$\Rightarrow y+\frac{18}{5}=\frac{48}{-46}(x-\frac{11}{5})$

$\Rightarrow y+\frac{18}{5}=\frac{24}{-23}x+\frac{24}{23}\times \frac{11}{5}$

$\Rightarrow y+\frac{24}{23}x+\frac{18}{5}-\frac{24}{23}\times \frac{11}{5}=0$

$\Rightarrow y+\frac{24}{23}x+\frac{1}{5}(18-\frac{264}{23})=0$

$\Rightarrow y+\frac{24}{23}x-\frac{1}{5}\left(\frac{414-209}{23}\right)=0$

$\Rightarrow y+\frac{24}{23}x-\frac{1}{5}\left(\frac{205}{23}\right)=0$

$\Rightarrow y+\frac{24}{23}x-\frac{41}{23}=0$

$\therefore 23y+24x-41=0$ is the required equation.