Question

find the area of the triangle whose vertices are (-8,4),(-6,6),(-3,9)

Answer 1

are of triangle=
1/2{ x1(y2-y3)+ x2(y3-y1) +x3(y1-y2)}

Answer 2

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

$\green{\tt{\therefore{Area\:of\:triangle=0\:sq\:units}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{: \implies Coordinate \: of \: A= (-8,4) } \\ \\ \tt{: \implies Coordinate \: of \: B = (-6,6) } \\ \\ \tt{: \implies Coordinate \: of \: C = (-3,9) } \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies Area \: of \: triangle = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} | x_{1} ( y_{2} - y_{3}) + x_{2}( y_{3} - y_{1}) + x_{3}( y_{1} - y_{2} ) | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |-8(6- 9) -6(9 - 4) - 3(4- 6)| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |-8\times -3 -6\times 5 - 3 \times -2 | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |24- 30 +6| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} \times 0} \\ \\ \green{\tt{: \implies Area \: of \: triangle =0\: sq \: units}} \\ \\ \purple{\bold{Some \: formula \: related \: to \: coordinate \: geometery}} \\ \pink{\tt{ \circ \: Distance \: formula = \sqrt{ (x_{2} - x_{1})^{2} + ( y_{2} - y_{1} )^{2} } }} \\ \\ \pink{\tt{ \circ \: Section \: formula = x= \frac{m x_{2} + n x_{1} }{m + n} }}$