Question
Find the area of $\triangle\text{ABC}$ whose vertices are:
A(1, 2), B(-2, 3) and C(-3, -4)

Answer

$A (1,2), B (-2,3)$ and $C (-3,-4)$ are the vertices of Then, $\left( x _1=1, y _1=2\right),\left( x _2=-2, y _2=3\right)$ and $\left( x _3=-3, y _3=-4\right)$ Area of triangle $A B C$
$=\frac{1}{2}\big\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big\}$
$=\frac{1}{2}\big\{1(3-(-4))+(-2))(-4-2)+(-3)(2-3)\big\}$
$=\frac{1}{2}\big\{1(3+4)-2(-6)-3(-1)\big\}$
$=\frac{1}{2}\big\{7+12+3\big\}$
$=\frac{1}{2}(22)$
$=11\ \text{sq}.\text{units}$

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