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Q. Area of the triangle with vertices $(- 2, 2), (1, 5) $and $(6, - 1)$ is

KEAMKEAM 2017Straight Lines

Solution:

Area of triangle having vertices $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by
Area $=\frac{1}{2}\begin{vmatrix}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{vmatrix}$
$\therefore $ Required area $=\frac{1}{2}\begin{vmatrix}-2 & 2 & 1 \\ 1 & 5 & 1 \\ 6 & -1 & 1\end{vmatrix}$
$ =\frac{1}{2}[-2(5+1)-2(1-6)+1(-1-30)] $
$=\frac{1}{2}[-12+10-31] $
$=\frac{-33}{2} $
$\therefore $ Area $=\frac{33}{2} $ sq units