Question
Find the area of $\triangle\text{ABC}$ whose vertices are:
$A(10, -6), B(2, 5)$ and $C(-1, 3)$
$A(10, -6), B(2, 5)$ and $C(-1, 3)$
✓
Answer
$A(10, -6), B(2, 5)$ and $C(-1, -3)$ are the vertices of $\triangle\text{ABC}.$
Then, $(x_1 = 10, y_1= -6), (x_2 = 2, y_2 = 5)$ and $(x_3 = -1, y_3 = 3)$
Area of triangle ABC
$=\frac{1}{2}\big\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big\}$
$=\frac{1}{2}\big\{10(5-3)+2(3-(-6))+(-1)(-6-5)\big\}$
$=\frac{1}{2}\big\{10(2)+2(9)-1(-11)\big\}$
$=\frac{1}{2}\big\{20+18+11\big\}$
$=\frac{1}{2}(49)$
$=24.5\ \text{sq}.\text{units}$
Then, $(x_1 = 10, y_1= -6), (x_2 = 2, y_2 = 5)$ and $(x_3 = -1, y_3 = 3)$
Area of triangle ABC
$=\frac{1}{2}\big\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big\}$
$=\frac{1}{2}\big\{10(5-3)+2(3-(-6))+(-1)(-6-5)\big\}$
$=\frac{1}{2}\big\{10(2)+2(9)-1(-11)\big\}$
$=\frac{1}{2}\big\{20+18+11\big\}$
$=\frac{1}{2}(49)$
$=24.5\ \text{sq}.\text{units}$
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