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

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

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

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

$\Rightarrow 7 x - y + 3 = \pm 5 (x + y - 3)$

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

$\Rightarrow x - 3 y + 9 = 0$ or $3 x + y - 3 = 0$

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

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

$\Rightarrow k = 31, L = - 7$

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

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