Question

The equation of the perpendicular bisectors of the sides AB and AC of a $\Delta$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. $2x+7y=-18$
• B. $14x-23y=40$
• C. $23x+14y=40$
• D. $23x-14y=40$

Answer 1

Answer: (A)

$2x+7y=-18$

We have,

The equation of perpendicular bisector of $AB$ is

$x-y+5=0$

$⟹y=x+5-\left(1\right)$

Slope $m=1$

The equation of $AB$ passing through the point $(1,-2)$ is-

$y-y_1=m(x-x_1)$

$⟹y+2=-1(x-1)$

$⟹y+2=-x+1$

$⟹x+y+1=0-\left(2\right)$

Given equation

$x+2y=0-\left(3\right)$

According to question

From equation (2) and (3) to,

From equation (2)

$x+y+1=0$

$⟹y=-(x+1)$

Put in (3) to,

$x+2y=0$

$⟹x-2(x+1)=0$

$⟹x-2x-2=0$

$⟹x=-2$

Put in equation (3)

$y=1$

Then,$(x_1,y_1)=(-2,1)$

From equation (1) and (3) to,

Point of intersection

$(x_2,y_2)=(-\frac{10}{3},\frac{5}{3})$

Then equation of

$y+2=\frac{\frac{5}{3}-1}{\frac{-10}{3}+2}(x+2)$

$⟹y+2=\frac{-2}{7}(x+2)$

$⟹7y+14=-2x-4$

$⟹2x+7y=-4-14$

$⟹2x+7y=-18$

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