Find the area of a triangle whose vertices are
$(6, 3), (- 3, 5)$ and $(4, - 2)$

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Solution

Vertices of the given triangle are $A (6, 3) B (- 3, 5), C (4, - 2)$

$∴$ Area of $△ A B C$

$= \frac{1}{2} ∣ y_{1} (x_{2} - x_{3}) + y_{2} (x_{3} - x_{1}) + y_{3} (x_{1} - x_{2}) ∣ s q . u n i t$

Here, $x_{1} = 6, y_{1} = 3, x_{2} = - 3, y_{2} = 5, x_{3} = 4, y_{3} = - 2$

$= \frac{1}{2} ∣ 3 (- 3 - 4) + 5 (4 - 6) + (- 2) (6 + 3) ∣ s q . u n i t$

$= \frac{1}{2} ∣ - 21 - 10 - 18 ∣ s q . u n i t$

$= \frac{1}{2} \times 49$ sq.units

$= \frac{49}{2}$ sq.units

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