Question

Let the points of intersections of the lines $x-y+1=0,x-2y+3=0$ and $2x-5y+11=0$ are the mid points of the sides of a triangle $ABC$. Then the area of the triangle $ABC$ is $\text{sq. units.}$

Answer 1

Let Point of intersection of lines are $D,E,F$
$\begin{array}{c}x-y+1=0\\ x-2y+3=0\\ x=1,y=2\end{array}|\begin{array}{c}x-y+1=0\\ 2x-5y+11=0\\ x=2,y=3\end{array}|\begin{array}{c}x-2y+3=0\\ 2x-5y+11=0\\ x=7,y=5\end{array}|$

$\text{Area of}\Delta ABC=4\cdot \text{(Area of}\Delta DEF)$
$\Delta ABC=4\times \frac{1}{2}\left|\begin{array}{ccc}1& 2& 1\\ 2& 3& 1\\ 7& 5& 1\end{array}\right|$
$=|2\left[1\right(3-5)+2(7-2)+1(10-21\left)\right]|$
$=|2\times [-2+10-11]|$
$=6$ sq. units