Question

answer the question please give me fast answer​

Answer 1

$Let \: A(2,3) = (x_{1} , y_{1}) , B(-1,0) = (x_{2} , y_{2})$

$and \: C(2,-4) = (x_{3} , y_{3})\:are \: Vertices$

$of \: a \: triangle \: ABC$

$Area \:of \: \triangle ABC$

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

$= \frac{1}{2}|2[0-(-4)] + (-1)(-4-3)+2(3-0)|$

$= \frac{1}{2}| 2\times 4 + (-1)(-7) + 6 |$

$= \frac{1}{2}| 8 + 7 + 6 |$

$= \frac{21}{2}$

$= 10.5 \: square \: units$

Therefore.,

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

$\green { = 10.5 \: square \: units}$

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Answer 2

Answer:

Area of the triangle enclosed by the given coordinates will be 10.5 sq. unit. The step by step solution for the same is in the attached photo.