Question
Find the area of $\triangle\text{ABC}$ whose vertices are:
$A(3, 8), B(-4, 2)$ and $C(5, -1)$
$A(3, 8), B(-4, 2)$ and $C(5, -1)$
✓
Answer
$A(3, 8), B(-4, 2)$ and $C(5, -1)$ are the vertices of $\triangle\text{ABC}.$
Then, $(x_1 = 3, y_1= 8), (x_2 = -4, y_2 = 2)$ and $(x_3 = 5, y_3 = -1)$
Area of triangle ABC
$=\frac{1}{2}\big\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big\}$
$=\frac{1}{2}\big\{3(2-(-1))+(-4)(-1-8)+5(8-2)\big\}$
$=\frac{1}{2}\big\{3(2+1)-4(-9)+5(6)\big\}$
$=\frac{1}{2}\big\{9+36+30\big\}$
$=\frac{1}{2}(75)$
$=37.5\ \text{sq}.\text{units}$
Then, $(x_1 = 3, y_1= 8), (x_2 = -4, y_2 = 2)$ and $(x_3 = 5, y_3 = -1)$
Area of triangle ABC
$=\frac{1}{2}\big\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big\}$
$=\frac{1}{2}\big\{3(2-(-1))+(-4)(-1-8)+5(8-2)\big\}$
$=\frac{1}{2}\big\{3(2+1)-4(-9)+5(6)\big\}$
$=\frac{1}{2}\big\{9+36+30\big\}$
$=\frac{1}{2}(75)$
$=37.5\ \text{sq}.\text{units}$
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