Question

Find the area of the triangle whose vertices are $(-5,-1),(3,-5)$and $(5,2)$

Answer 1

Let the given points be $A(-5,-1),B(3,-5)$ and $C(5,2).$

Here, we have

$x_1=-5,y_1=-1$

$x_2=3,y_2=-5$

and $x_3=5,y_3=2$

Now, Area of $\Delta ABC$

$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(v_1-y_2)]$

$=\frac{1}{2}\left[\right(-5\left)\right\{-5-2\}+(3\left)\right\{2-(-1)\}+(5\left)\right\{(-1)-(-5)\left\}\right]$

$=\frac{1}{2}[35+9+20]=32sq$ units