Question

Sol. [Key Concept: The base makes equal angles with the two equal sides (slopes $7, - 1$); solving gives base slopes $\frac{13}{}$ and $- 3$ through $(1, - 10)$.]

$3^{\text{rd}\text{~}}$ side will be parallel to angle bisectors of given lines,

$$\begin{matrix} & \frac{x - y +}{5\sqrt{2}} = \pm \frac{x + y -}{\sqrt{2}} \\ \Rightarrow \ 2x - 6y & \ = c \end{matrix}$$

$\ \ 3^{\text{rd}\text{~}}$ side will be given by

$$x - 3y = 31\text{~}\text{or}\text{~}3x + y = - 7$$

✔ (a) $x - 3y - 31 = 0$ (slope $\frac{13}{}$) through $(1, - 10)$.

✘ (b) $y - 3x + 13 = 0$ (slope 3) is not equally inclined to the equal sides.

✔ (c) $3x + y + 7 = 0$ (slope $- 3$) through $(1, - 10)$.

✘ (d) $y - 2x + 12 = 0$ (slope 2) is not a valid base.