The equations of two equal sides AB and AC of an isosceles triangle ABC are $x + y = 5$ and $7x–y=3$, respectively, and the area $\left(\mathrm{△}ABC\right)=5{\text{ unit }}^2$ then straight line join $\overline{BC}$ can be

Side BC will be angle bisector to the bisector of $\angle BAC$ Now, the equations of the bisectors of lines AB and AC are $\frac{(x+y-5)}{\sqrt{2}}=\pm \frac{(7x-y-3)}{5\sqrt{2}}$

Then the equation of AD is 3x + y – 7 = 0.

If AD=$\lambda$  , then

$\mathrm{\Delta }ABC=\frac{1}{2}\lambda \left|BC\right|=\frac{1}{2}\times \lambda \times 2\lambda |\tan \theta |={\lambda }^2|\tan \theta |$