Math, asked by shreesha31, 1 year ago

find the area of the triangle whose vertices are (0,0),(5,3) and (1,9)​


shreesha31: please help

Answers

Answered by vanaparthibhavani70
1

the area of the triangle (0,0) (5,3) (1,9) is 21

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shreesha31: thanks bro for helping me
vanaparthibhavani70: its ok sister
Answered by BrainlyConqueror0901
5

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

\green{\tt{\therefore{Area\:of\:triangle=21\:sq\:units}}}

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

 \green{ \underline \bold{Given : }} \\  \tt{: \implies Coordinate \: of \: A= (0,0) } \\  \\ \tt{: \implies Coordinate \: of \: B = (5,3) } \\  \\ \tt{: \implies Coordinate \: of \: C = (1,9) } \\  \\ \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}  |0(3- 9) + 5(9 - 0) + 1(0 - 3)| } \\  \\ \tt{:  \implies Area \: of \: triangle = \frac{1}{2}  |0 \times -6 +  5\times 9 + 1 \times -3 | } \\  \\ \tt{:  \implies Area \: of \: triangle = \frac{1}{2}  |0 + 45-3| } \\  \\ \tt{:  \implies Area \: of \: triangle = \frac{1}{2} \times 42} \\  \\   \green{\tt{:  \implies Area \: of \: triangle =21 \: 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} }}

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