Question

The equation of the perpendicular bisector of the sides

$AB$ and $AC$ of triangle $ABC$ are $x-y+5=0$ and $x+2y=0$ respectively. If the point $A$ is $(1,-2)$, the equation of the line $BC$ is

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

Answer 1

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

(a) Given equation of line $LF$ is $x-y+5=0$ ...(1)

and equation of line $ME$ is $x+2y=0$ ...(2)

Since $AB$ is perpendicular to $LF$ and passes throught $A$

$(1,-2)$ therefore equation of $AB$ is $x+y-(1-2)=0$ $\Rightarrow x+y=-1$ ...(3)

Similarly, equation of line $AC$ is $2x-y-(2.1+2)=0\Rightarrow 2x-y=4$ ...(4)

Solving (1) and (3), we get $x=-3,y=2$

$\therefore F\equiv (-3,2)$

Solving (2) and (4) we get $x=8/5,y=-4/5$

$\therefore E\equiv (8/5,-4/5)$

Let $B\equiv (x_1,y_1)$ and $C\equiv (x_2,y_2)$

Since $F$ is the middle point of $AB$

$\therefore \frac{x_1+1}{2}=-3\Rightarrow x_1=-7$ and $\frac{y_1-2}{2}=2\Rightarrow y_1=6$

Hence $B\equiv (-7,6).$

Again $E$ is the middle point of $AC$

$\therefore \frac{8}{5}=\frac{x_2+1}{2}$

$\Rightarrow x_2=\frac{11}{5}$ and $-\frac{4}{5}=\frac{y_2-2}{2}$

$\Rightarrow y_2=\frac{2}{5}$

Hence $C\equiv (\frac{11}{5},\frac{2}{5})$

$\Rightarrow$ equation of $BC$ is $14x+23y=40$