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The area of a triangle having vertices $A (3, 2), B (11, 8)$ and $C (8, 12)$ is

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answer is A.

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Detailed Solution

$A (x_{1}, y_{1}) = (3, 2)$

$B (x_{2}, y_{2}) = (11, 8)$

$C (x_{3}, y_{3}) = (8, 12)$

Area $= \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | 3 (8 - 12) + 11 (12 - 2) + 8 (2 - 8) |$

$= \frac{1}{2} | - 12 + 110 - 48 | = \frac{1}{2} | 50 | = 25 sq.units$

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