Question

Find the area of triangle whose vertices are( 1,3), (-7,6) ,(5,-1)

Answer 1

★★★★★★★★★★★

here,
x1=1,y1=3

x2=-7,y2=6

x3=5,y3=-1



☆Area of triangle:


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

=>1/2{1(6+1)-7(-1-3)+5(3-6)}

=>1/2{7+28-15)

=>1/2×20

=>10

Area=10cm^2

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●hope it helps

Answer 2

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

$\green{\tt{\therefore{Area\:of\:triangle=10\:sq\:units}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{: \implies Coordinate \: of \: A= (1,3) } \\ \\ \tt{: \implies Coordinate \: of \: B = (-7,6) } \\ \\ \tt{: \implies Coordinate \: of \: C = (5,-1) } \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies Area \: of \: triangle = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} | x_{1} ( y_{2} - y_{3}) + x_{2}( y_{3} - y_{1}) + x_{3}( y_{1} - y_{2} ) | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |1(6 - (-1)) -7(-1 - 3) + 5(3 - 6)| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |1\times 7 -7\times -4 + 5 \times -3 | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |7 + 28 -15| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} \times 20} \\ \\ \green{\tt{: \implies Area \: of \: triangle =10\: sq \: units}} \\ \\ \purple{\bold{Some \: formula \: related \: to \: coordinate \: geometery}} \\ \pink{\tt{ \circ \: Distance \: formula = \sqrt{ (x_{2} - x_{1})^{2} + ( y_{2} - y_{1} )^{2} } }} \\ \\ \pink{\tt{ \circ \: Section \: formula = x= \frac{m x_{2} + n x_{1} }{m + n} }}$