Question

find the area of triangle whose vertices is A(2,3),B(-2,1)and C(3,-2)
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Answer 1

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

$\:B(-2,1) = (x_{2} , y_{2}) \: and$

$C(3,-2) = (x_{3} , y_{3})$

$Area \: of \: triangle \: ABC$

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

$= \frac{1}{2}|2[1-(-2)]+(-2)(-2-3)+3(3-1)|$

$= \frac{1}{2}|2(1+2)+(-2)(-5)+3\times 2|$

$= \frac{1}{2}|2\times 3 + 10 + 6 |$

$= \frac{1}{2}|6+10+6|$

$= \frac{1}{2} \times 22$

$= 11 \: square \: units$

Therefore.,

$\red{Area \: of \: triangle \: ABC }$

$\green{ = 11 \: square \: units}$

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