Question

Determine, the area of the triangle formed by the lines:
$3x-y=3$
$2x-3y=2$
$x+2y=8$

Answer 1

Given lines

$3x-y=3⟶\left(1\right)$

$2x-3y=2⟶\left(2\right)$

$x-2y=8⟶\left(3\right)$

By solving $\left(1\right)$ and $\left(3\right)$

$3x-y=3,x+2y=8$

$6x-2y=6,x+2y=8$

$7x=14⟹x=2,y=3$

Solving $\left(2\right)$ and $\left(3\right)$

$2x-3y=2,2x+4y=16$

$7y=14⟹y=2,x=4$

Area of $△ABC$=Area of $△ABD$-Area of $△ACD$

$=\frac{1}{2}\left[\left(3\right)\left(7\right)\right]-\frac{1}{2}\times \left(2\right)\times \left(7\right)=\frac{7}{2}$