Question

Find the area of the triangle whose sides are along the lines $x-y=-1,x+y=5$ and $x-3y=-3$

Answer 1

Given,
$x-y=-1...\left(1\right)$
$x+y=5...\left(2\right)$
$x-3y=-3...\left(3\right)$
Solving eqn(1) and (2), we get
$x=2,y=3$
Solving eqn(1) and (3), we get
$x=0,y=1$
Solving eqn(2) and (3), we get
$x=3,y=2$
Therefore, vertices of the trianle are $(2,3),(0,1)$ and $(3,2)$

Required area is given as,
$\Delta =\frac{1}{2}\left|\begin{array}{ccc}2& 3& 1\\ 0& 1& 1\\ 3& 2& 1\end{array}\right|$
$=\frac{1}{2}\times 4$
$=2\text{sq. units}$