Question
Find the area of a triangle whose vertices are,
(a, c + a), (a, c) and (-a, c - a).
(a, c + a), (a, c) and (-a, c - a).
✓
Answer
Coordinates of $\triangle\text{ABC}$ are (a, c + a), (a, c) and (-a, c - a)
$\therefore$ Area of $\triangle\text{ABC}=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[\text{a}(\text{c}-\text{c}+\text{a})+\text{a}(\text{c}-\text{a}-\text{c}-\text{a})+(-\text{a})(\text{c}+\text{a}-\text{c})]$
$=\frac{1}{2}(\text{a}^2-2\text{a}^2-\text{a}^2)$
$=\frac{1}{2}(-2\text{a}^2)$
$=\text{a}^2\text{ sq.units}$
$\therefore$ Area of $\triangle\text{ABC}=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$=\frac{1}{2}[\text{a}(\text{c}-\text{c}+\text{a})+\text{a}(\text{c}-\text{a}-\text{c}-\text{a})+(-\text{a})(\text{c}+\text{a}-\text{c})]$
$=\frac{1}{2}(\text{a}^2-2\text{a}^2-\text{a}^2)$
$=\frac{1}{2}(-2\text{a}^2)$
$=\text{a}^2\text{ sq.units}$
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