Find the area of triangle having vertices $A (6, 10), B (4, 5)$ and $C (3, 8)$ .

5

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5.5

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10

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11

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Solution

The correct option is C $5.5$
The point are $A (x_{1}, y_{1}) = (6, 10), B (x_{2}, y_{2}) = (4, 5)$ and $C (x_{3}, y_{3}) = (3, 8)$ .
$a r Δ A B C = \frac{1}{2} {x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})}$
$\Rightarrow a r Δ A B C = \frac{1}{2} {6 (5 - 8) + 4 (8 - 10) + 3 (10 - 5)}$ sq. units
$\Rightarrow a r Δ A B C = 5.5$ sq. units

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