Question

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Find the area of the triangle whose coordinates of vertices are A(6,3), B(-3,5) and (4,-2).


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Answer 1

$\large{\blue{\bold{Formula \: Used}}} \: \\ \scriptsize{\bf \: Area \: of \: triangle = \frac{1}{2} \times x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})} \\ \\ \large{\purple{\bold{\underline{Step \: by \: step \: Solution:}}}} \: \\ \small\tt \:{ Let \: us \: consider \: the \: respective \: coordinates \: as : } \\ \bf \: A(6, 3) \: \rightarrow \: \: (x_{1}, y_{1}) \\ \bf \: B(-3,5) \rightarrow \: \: (x_{2}, y_{2}) \\ \bf \: C(4,-2) \rightarrow \: \: (x_{3}, y_{3}) \\ \\ \large \bold{Calculation }\\ \sf \small{ \: Area \: of \: triangle = \frac{1}{2} \times 6(5 - ( - 2)) - 3( - 2 - 3) + 4(3 - 5)} \\ \\ \small{\sf \: Area \: of \: triangle = \frac{1}{2} \times( 6 \times 7) +(3 \times 5) - (4 \times 2) }\\ \\ \sf \: Area \: of \: triangle = \frac{1}{2}\times( 42 + 15 - 8) \\ \\ \sf \: Area \: of \: triangle = \frac{1}{2} \times 49 \\ \\ \sf \: Area \: of \: triangle = 24.5 \: \\ \\ \\ \large\boxed{ \sf \: The \: area \: of \: triangle \: is \: 24.5 \: Square \: units.}$

Answer 2

Answer:

area=24.5 sq is right answer