Question

Two equal sides of an isosceles triangle are given by the equations $7x-y+3=0$ and $x+y-3=0$ and its third side passes through the point $(1,-10)$. Determine the equation of the third side.

Answer 1

Since, $\Delta$ is isosceles triangle so, third side is equally inclined to the lines $7x-y+3=0$ and $x+y-3=0$

Hence, third line is parallel to their angle in sector.

$\frac{7x-y+3}{\sqrt{50}}=\pm \frac{x+y-3}{\sqrt{2}}$

$3x+y-3=0$ -----(i)

$x-3y+9=0$ -----(ii)

The line through $(1,-10)$ parallel to $\left(i\right)$ is $3x+y+7=0$ and that parallel to $\left(ii\right)$ is $x-3y-31=0.$

Which is the required answer.