Question

Two equal sides of an isosceles triangle have the equations $7x-y+3=0$and $x+y-3=0$ and its third side passes through the point $\left(1,-10\right)$. Find the equation of third side.

Answer 1

Since the Sides of the isosceles triangle has two sides equal so the angle opposite to these lines are also equal which

they will make with the remaining side

Let the Slope of other line be $m$

and as we can see that slope of the equal sides are $7,-1$

Now, as the opposite angles are equal

$\tan {\theta }_1=\tan {\theta }_2$

$\Rightarrow |\frac{7-m}{1+7m}|=|\frac{m-(-1)}{1+(-m)}|$

$\Rightarrow |\frac{7-m}{1+7m}|=|\frac{m+1}{1-m}|$

$\Rightarrow 6m^2+16m-6=0$

$\Rightarrow 2(3m-1)(m+3)=0\Rightarrow m=\frac{1}{3},-3$

Hence, Equation of other line will be $y+10=m(x-1)$

$\Rightarrow y+10=\frac{1}{3}(x-1)\Rightarrow 3y-x+31=0$

$\Rightarrow y+10=-3(x-1)\Rightarrow 3x+y+7=0$

Hence, above are the 2 possible lines