Question

In triangle $ABC$, equation of the right bisectors of the sides $AB$ and $AC$ are $x=0$ and $y=0$ respectively. If $A\equiv (5,7)$ then equation of side $BC$ is

• A. $7x+5y=0$
• B. $7x-5y=0$
• C. $7x+1y=0$
• D. $Noneofthese$

Answer 1

The correct option is A $7x+5y=0$

$AB\perp x-axis$

$AC\perp y-axis$

Let $B=(x_1,y_1)$

$C=(x_2,y_2)$

$AB\perp x-axis$

Slope of $AB=\infty$

$\Rightarrow x_1=5$

$\therefore B=(5,y_1)$

$AC\perp y-axis$

$\Rightarrow$ slope of $AC=0$

$\Rightarrow y_2=7$

$C=(x_2,7)$

For right angled $△",\perp$ bisectors of perpendicular sides meet at mid point of hypotenuse,

$\Rightarrow \frac{5+x_2}{2}=0,\frac{7+y_1}{2}=0$

$x_2=-5,y=-7$

$\therefore$ Equation of $BC=(y+y)=\frac{14}{-10}(x-5)$

$\Rightarrow -10y-70=14x-70$

$\Rightarrow 7x+5y=0$

Hence, the answer is $7x+5y=0.$