Question
Find the area of $\triangle\text{ABC}$ whose vertices are:
$A(-5, 2), B(-4, -5)$ and $C(4, 5)$
$A(-5, 2), B(-4, -5)$ and $C(4, 5)$
✓
Answer
$A(-5, 7), B(-4, -5)$ and $C(4, 5)$ are the vertices of $\triangle\text{ABC}.$
Then, $(x_1 = -5, y_1= 7), (x_2 = -4, y_2 = -5)$ and $(x_3 = 4, y_3 = 5)$
Area of triangle ABC
$=\frac{1}{2}\big\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big\}$
$=\frac{1}{2}\big\{(-5)(-5-5)+(-4)+(5-7)\\+(-4-2)+(4)(7-(-5))\big\}$
$=\frac{1}{2}\big\{(-5)(-10)-4(-2)+4(12)\big\}$
$=\frac{1}{2}\big\{50+8+48\big\}$
$=\frac{1}{2}(106)$
$=53\ \text{sq}.\text{units}$
Then, $(x_1 = -5, y_1= 7), (x_2 = -4, y_2 = -5)$ and $(x_3 = 4, y_3 = 5)$
Area of triangle ABC
$=\frac{1}{2}\big\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\big\}$
$=\frac{1}{2}\big\{(-5)(-5-5)+(-4)+(5-7)\\+(-4-2)+(4)(7-(-5))\big\}$
$=\frac{1}{2}\big\{(-5)(-10)-4(-2)+4(12)\big\}$
$=\frac{1}{2}\big\{50+8+48\big\}$
$=\frac{1}{2}(106)$
$=53\ \text{sq}.\text{units}$
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