Question

Find the area of the triangle formed by the straight lines whose equations are
$y=x,y=2x$, and $y=3x+4$.

Answer 1

$y=x.....\left(i\right)y=2x......\left(ii\right)y=3x+\mathrm{4......}\left(iii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow x=0,y=0$

So the point of intersection is $A(0,0)$

solving $\left(ii\right)$ and $\left(iii\right)$

$\Rightarrow x=-4,y=-8$

so the second point of intersection is $B(-4,-8)$

solving $\left(i\right)$ and $\left(iii\right)$

$\Rightarrow x=-2,y=-2$

so the third point of intersection is $C(-2,-2)$

So the vertices of triangle are $A(0,0),B(-4,-8)$ and $(-2,-2)$

Area of triangle $=△=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ -4& -8& 1\\ -2& -2& 1\end{array}\right|$

$△=\frac{1}{2}\left\{0\right(-8\times 1-1\times -2)-0(-4\times 1-1\times -2)+1(-4\times -2-)-(-4)\times -8)\}△=\frac{1}{2}|-8|△=4$