Question:
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
Solution:
intersection point of give lines are $(1,2),(7,5)$, $(2,3)$
$\Delta=\frac{1}{2}\left|\begin{array}{lll}1 & 2 & 1 \\ 7 & 5 & 1 \\ 2 & 3 & 1\end{array}\right|$
$=\frac{1}{2}[1(5-3)-2(7-2)+1(21-10)]$
$=\frac{1}{2}[2-10+11]$
$\Delta \mathrm{DEF}=\frac{1}{2}(3)=\frac{3}{2}$
$\Delta \mathrm{ABC}=4 \Delta \mathrm{DEF}=4\left(\frac{3}{2}\right)=6$
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