Find the area of a triangle whose vertices are A (3, 2), B (11, 8) and C(8, 12). [3 MARKS]

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Solution

Concept : 1 Mark
Application : 1 Mark
Calculation : 1 Mark

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

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

and $C = (x_{3}, y_{3}) = (8, 12)$ be the given points. Then,

$A r e a o f Δ A B C = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$A r e a o f Δ A B C = \frac{1}{2} | 3 (8 - 12) + 11 (12 - 2) + 8 (2 - 8) |$

$A r e a o f Δ A B C = \frac{1}{2} | - 12 + 110 - 48 | = 25 s q . u n i t s$

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