Let the points of intersections


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


intersection point of give lines are $(1,2),(7,5)$, $(2,3)$

$\Delta=\frac{1}{2}\left|\begin{array}{lll}1 & 2 & 1 \\ 7 & 5 & 1 \\ 2 & 3 & 1\end{array}\right|$



$\Delta \mathrm{DEF}=\frac{1}{2}(3)=\frac{3}{2}$

$\Delta \mathrm{ABC}=4 \Delta \mathrm{DEF}=4\left(\frac{3}{2}\right)=6$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now