Two equal sides of an isosceles triangle are 7x - y + 3 = 0 and x +...

The third side is parallel to a bisector of the angle between equal sides.

The bisectors are $\dfrac{7x - y + 3}{{7}^{2} + \left(- 1\right)^{2}} = \pm \dfrac{( x + y - 3 )}{{1}^{2} + {1}^{2}}$

$⇒ \dfrac{7x - y + 3}{5\sqrt{2}} = \pm \dfrac{( x + y - 3 )}{2}$

$⇒ 7x - y + 3 = \pm 5 ( x + y - 3 )$

$⇒ 2x - 6y + 18 = 0$ or $12x + 4y - 12 = 0$

$⇒ x - 3y + 9 = 0$ or $3x + y - 3 = 0$

Let the third side be $x - 3y = k$ or $3x + y = L$

It passes through $( 1, - 10 )$.

$⇒ k = 31, L = - 7$

Hence, required lines are $x - 3y = 31$ or $3x + y = - 7$.&