Find the area of the triangle, the equations whose sides are $\;$ $y + x = 0$, $\;$ $y = x + 6$ $\;$ and $\;$ $y = 7x + 5$.
The sides of the triangle are
$x + y = 0$ $\;\;\; \cdots \; (1)$,
$y = x + 6$ $\;$ i.e. $\;$ $x - y + 6 = 0$ $\;\;\; \cdots \; (2)$,
$y = 7x + 5$ $\;$ i.e. $\;$ $7x - y + 5 = 0$ $\;\;\; \cdots \; (3)$
Let equations $(1)$ and $(2)$ intersect at point $A$.
Let equations $(2)$ and $(3)$ intersect at point $B$.
Let equations $(3)$ and $(1)$ intersect at point $C$.
Solving equations $(1)$ and $(2)$ simultaneously gives the point of intersection as $\;$ $A \left(x_1, y_1\right) = \left(-3, 3\right)$
Solving equations $(2)$ and $(3)$ simultaneously gives the point of intersection as $\;$ $B \left(x_2, y_2\right) = \left(\dfrac{1}{6}, \dfrac{37}{6}\right)$
Solving equations $(3)$ and $(1)$ simultaneously gives the point of intersection as $\;$ $C \left(x_3, y_3\right) = \left(\dfrac{-5}{8}, \dfrac{5}{8}\right)$
Let $AD$ be the altitude drawn from point $A$ to line given by equation $(3)$.
Then, length of altitude $AD = \left|\dfrac{7 \times \left(-3\right) -1 \times 3 + 5}{\sqrt{7^2 + \left(-1\right)^2}}\right| = \dfrac{19}{\sqrt{50}} = \dfrac{19}{5 \sqrt{2}}$ units
Distance $BC = \sqrt{\left(\dfrac{1}{6}+ \dfrac{5}{8}\right)^2 + \left(\dfrac{37}{6} - \dfrac{5}{8}\right)^2} = \sqrt{\left(\dfrac{19}{24}\right)^2 + \left(\dfrac{133}{24}\right)^2} = \dfrac{\sqrt{18050}}{24} = \dfrac{95 \sqrt{2}}{24}$ units
$\therefore \;$ Area of triangle whose sides are the given equations is
$= \dfrac{1}{2} \times AD \times BC = \dfrac{1}{2} \times \dfrac{19}{5 \sqrt{2}} \times \dfrac{95 \sqrt{2}}{24} = \dfrac{361}{48}$ sq units