Question

Find the area of the triangle formed by the straight lines whose equations are
$2y+x-5=0,y+2x-7=0$, and $x-y+1=0$.

Answer 1

$2y+x-5=\mathrm{0.....}\left(i\right)y+2x-7=\mathrm{0.......}\left(ii\right)x-y+1=\mathrm{0.........}\left(iii\right)$

solving $\left(i\right)$ and $\left(ii\right)$ by substituting $y$ from $\left(ii\right)$ in $\left(i\right)$

$2(7-2x)+x-5=014-4x+x-5=03x=9\Rightarrow x=3y+2\left(3\right)-7=0\Rightarrow y=1$

so their point of intersection is $A(3,1)$

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

$\Rightarrow x=2,y=3$

so the second point of intersection is $B(2,3)$

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

$\Rightarrow x=1,y=2$

so the third point of intersection is $C(1,2)$

If the vertices of triangle are $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ then

Area of triangle $=△=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Here, the vertices of triangle are $A(3,1),B(2,3)$ and $C(2,1)$

Area of triangle $=△=\frac{1}{2}\left|\begin{array}{ccc}3& 1& 1\\ 2& 3& 1\\ 1& 2& 1\end{array}\right|$

$△=\frac{1}{2}\left\{3\right(3\times 1-1\times 2)-1(2\times 1-1\times 1)+1(2\times 2-3\times 1)△=\frac{1}{2}\left|3\right|△=\frac{3}{2}$