Question

Find the area of the triangle formed by the straight lines whose equations are
$y+x=0,y=x+6$, and $y=7x+5$.

Answer 1

$y+x=\mathrm{0......}\left(i\right)y=x+\mathrm{6.......}\left(ii\right)y=7x+\mathrm{5.......}\left(ii\right)$

solving $\left(i\right)$ and $\left(ii\right)$ by substituitng $y$ form $\left(ii\right)$ in $\left(i\right)$

$x+x+6=0\Rightarrow x=-3y=x+6\Rightarrow y=3$

So their point of intersection is $A(-3,3)$

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

$\Rightarrow x=\frac{1}{6},y=\frac{37}{6}$

So second point of intersection is $B(\frac{1}{6},\frac{37}{6})$

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

$\Rightarrow x=-\frac{5}{8},y=\frac{5}{8}$

So the third point of intersection is $C(-\frac{5}{8},\frac{5}{8})$

So the vertices of triangle are $A(-3,3),B(\frac{1}{6},\frac{37}{6})$ and $C(-\frac{5}{8},\frac{5}{8})$

Area of triangle $=\frac{1}{2}\left|\begin{array}{ccc}-3& 3& 1\\ \frac{1}{6}& \frac{37}{6}& 1\\ -\frac{5}{8}& \frac{5}{8}& 1\end{array}\right|$

$=\frac{1}{2}\{-3(\frac{37}{6}\times 1-1\times \frac{5}{8})-3(\frac{1}{6}\times 1-1\times -\frac{5}{8})+1(\frac{1}{6}\times \frac{5}{8}-\frac{37}{6}\times -\frac{5}{8})\}=\frac{1}{2}|-\frac{361}{24}|=\frac{361}{48}=7\frac{25}{48}$