Question

The equation 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. $14x+23y=40$
• B. $14x-23y=40$
• C. $23x+14y=40$
• D. $23x-14y=40$

Answer 1

The correct option is A $14x+23y=40$

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

Since coordinates of A are $(1,-2)$,

$\therefore$ the slope of $AB=\frac{\beta +2}{\alpha -1}$ ...(1)

The equation of the perpendicular bisector of $AB$ is $x-y+5=0$ ...(2)

From(1) and (2), we have $\left(\frac{\beta +2}{\alpha -1}\right)=-1$

$\Rightarrow \alpha +\beta +1=0$ ...(3)

Also, the mid point of $AB$ lies on (2),

$\therefore \left(\frac{\alpha +1}{2}\right)-\left(\frac{\beta -2}{2}\right)+5=0$

$\Rightarrow \alpha -\beta +13=0$ ...(4)

Solving (3) and (4), we get $\alpha =-7$ and $\beta =6.$

So, the coordinates of $B$ are $(-7,6).$

Similarly, the coordinatrs of $C$ are $(\frac{11}{5},\frac{2}{5}).$

$\therefore$ the equation of the line $BC$ is $y-6=\frac{\frac{2}{5}-6}{\frac{11}{5+7}}(x++7)$

$\Rightarrow y-6=-\frac{28}{46}(x+7)\Rightarrow 23(y-6)+14(x+7=0$

$\Rightarrow 14x+23y=40$

Hence,the equation of the line$BC$ is $\Rightarrow 14x+23y=40.$