Question

The equations of the perpendicular bisectors of the sides $AB$ and $AC$ of a triangle $ABC$ are $x-y+5=0$ and $x+2y=0$ respectively. If the co-ordinates of $A$ are $(1,-2)$, then the equation of $BC$ is

• A. $23x+14y-40=0$
• B. $14x+23y-40=0$
• C. $23x-14y+40=0$
• D. None of these

Answer 1

The correct option is B $14x+23y-40=0$

$\because AB$ line is perpendicular to $EO$(perpendicular bisector to side $AB$) and passes through vertex $A$, Equation of $AB::x+y+1=0$
Coordinates of $E=(\frac{x_1+1}{2},\frac{y_1-2}{2})$
Which is also intersection point of $AB$ with $x-y+5=0$(EO line equation)
So,
$\frac{x_1+1}{2}=-3;\frac{y_1-2}{2}=2$
$\Rightarrow (x_1,y_1)\equiv (-7,6)$
similarly equation of $AC::y=2x-4$
and coordinates of $F=(\frac{x_2+1}{2},\frac{y_2-2}{2})$
Which is also intersection point of $AC$ with $x+2y=0$(F0 line equation)
So,
$\frac{x_2+1}{2}=\frac{8}{5};\frac{y_2-2}{2}=-\frac{4}{5}$
$\Rightarrow (x_2,y_2)=(\frac{11}{5},\frac{2}{5})$
So, the equation of side$BC:$
$y-6=\frac{\frac{2}{5}-6}{\frac{11}{5}+7}(x+7)$

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

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

$\Rightarrow 23y-138=-14x-98$

$\Rightarrow 23y-138+14x+98=0$

$\Rightarrow 14x+23y-40=0$