Question

find the area of the triangle whose vertics are (4,7)(1,3)(5,1)

Answer 1

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Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Area\:of\:triangle=11\:sq\:units}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ \tt{: \implies Coordinate \: of \: A= (4,7) } \\ \\ \tt{: \implies Coordinate \: of \: B = (1,3) } \\ \\ \tt{: \implies Coordinate \: of \: C = (5,1) } \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies Area \: of \: triangle = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} | x_{1} ( y_{2} - y_{3}) + x_{2}( y_{3} - y_{1}) + x_{3}( y_{1} - y_{2} ) | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |4(3- 1) + 1(1 - 7) + 5(7 - 3)| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |4 \times 2+ 1\times -6 + 5 \times 4 | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |8 -6 +20| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} \times 22} \\ \\ \green{\tt{: \implies Area \: of \: triangle =11 \: sq \: units}} \\ \\ \purple{\bold{Some \: formula \: related \: to \: coordinate \: geometery}} \\ \pink{\tt{ \circ \: Distance \: formula = \sqrt{ (x_{2} - x_{1})^{2} + ( y_{2} - y_{1} )^{2} } }} \\ \\ \pink{\tt{ \circ \: Section \: formula = x= \frac{m x_{2} + n x_{1} }{m + n} }}$