Question

Find the area of a triangle whose three sides are having the equations $x+y=2,x-y=0$ and $x+2y-6=0$.

Answer 1

The given equations are
$x+y=2$ .....(1)
$x-y=0$ ......(2)
and $x+2y-6=0$ ......(3)
Solving (1) and (3), we get
$x+y=2$ .......(1)
$x+2y=6$ .....(3)
------------------------
$-y=-4$
Therefore, $y=4$
Solving (1) and (2), we get
$x+y=2$ .....(1)
$x-y=0$ ......(2)
-----------------
$2x=2$
$\Rightarrow x=1$
Substituting $y=4$ in (1), we get
$x+4=2$
$\Rightarrow x=-2$
Thus, $A(-2,4)$
Substituting $x=1$ in (1), we get
$1+y=2$
$\Rightarrow y=1$
Thus, $B(1,1)$
Solving (2) and (3), we get
$x-y=0$ ......(2)
$x+2y=6$ .......(3)
----------------------------
$-3y=-6$
Therefore, $y=2$
Substituting $y=2$ in (2), we get
$x-2=0$
$\Rightarrow x=2$
Thus, $C(2,2)$
Area of $△ABC=\frac{1}{2}9-2+2+8-(4+2-4)]$
$=\frac{1}{2}(8-2)=3$ sq.cm