Question

The area of ∆ABC with vertices A(1, −1), B(−4, 6) and C(−3, −5) is

(a) 52 sq units
(b) 27 sq units
(c) 48 sq units
(d) 24 sq units

Answer 1

(d) 24 sq units
The given points are $A\left(1,-1\right),B\left(-4,6\right)\text{and}C\left(-3,-5\right)$.
Here, $\left(x_1=1,y_1=-1\right),\left(x_2=-4,y_2=6\right)\text{and}\left(x_3=-3,y_3=-5\right)$
Therefore,
$\text{Area of}∆ABC=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$
$$\begin{gathered} =\frac{1}{2}\left[1\left(6+5\right)+\left(-4\right)\left(-5+1\right)+\left(-3\right)\left(-1-6\right)\right] \\ =\frac{1}{2}\left[11+16+21\right] \\ =\left(\frac{1}{2}\times 48\right) \\ =24\text{sq units} \end{gathered}$$