Explanation

(a)
The vertices of triangle are the intersection points of the lines
$x + y = 0\$…(i)
$3x + y = 4$ …(ii)
and $x + 3y = 4$ ...(iii)
On solving Eqs. (i) and (ii), (ii) and (iii), (iii) and (i), we get, the vertices of triangle are $A( - 2,\ 2),\ B(1,\ 1)$ and $C(2,\ - 2)$
Now, $AB = \sqrt{(1 + 2)^{2} + (1 - 2)^{2}}$
$= \sqrt{9 + 1} = \sqrt{10}$
$BC = \sqrt{(2 - 1)^{2} + ( - 2 - 1)^{2}}$
$= \sqrt{1^{2} + ({- 3)}^{2}} = \sqrt{10}$
$AC = \sqrt{(2 + 2)^{2} + ( - 2 - {2)}^{2}}$
$= \sqrt{16 + 16} = 4\sqrt{2}$
$\because AB = BC$
$\therefore$ Triangle is isosceles