Question

find the area of triangle whose vertices are A (3,2) B(11,8) C (8,12).​

Answer 1

Step-by-step explanation:

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Answer 2

$Let \: A(3,2) = (x_{1},y_{1}) , \: B(11,8) = (x_{2},y_{2}) \: and \: C(8,12) = (x_{3},y_{3}) \:are \\verticies \: of \: a \: triangle \: ABC$

$Area \:of \: the \: \triangle ABC \\= \frac{1}{2}|x_{1}(y_{2}-y_{3}) +x_{2}(y_{3}-y_{1}) +x_{3}(y_{2}-y_{1}) | \\= \frac{1}{2}|3(8-12)+11(12-2)+8(2-8)|\\= \frac{1}{2}| 3(-4)+ 11\times 10+8(-6)| \\= \frac{1}{2}|-12+110-48|\\= \frac{1}{2}|110 - 60| \\= \frac{1}{2} \times 50 \\= 25 \: square \:units$

Therefore.,

$\red{Area \:of \: the \: \triangle ABC}\\\green {=25 \: square \:units}$

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