Question

find the area of triangles whose vertices are (2,0),(-1,3)and(2,4)?

Answer 1

Heya user..!!

Here is ur answer..!!

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Answer:

By using the formula of area of triangle then we get the area of triangles whose vertices are (2,0),(-1,3)and(2,4).

Lets start solving..!!

x₁ =       2                 ,   y₁ = 0

x₂ =      -1                 ,  y₂ = 3

x₃ =        2                 , y₃ = 4

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

1/2 { 2 ( 3 - 4 ) - 1 (4 - 0 )+ 2 ( 0 - 3 ) }

1/2 ( 2 ( - 1 ) - 4 - 6 )

1/2 ( - 2 - 10 )

1/2 ( -12 )

→ - 6

So here minus sign is neglected.

∴ Area of triangles whose vertices are (2,0),(-1,3)and(2,4) are : 6cm.

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I Hope this may help u..!!

Be Brainly..!!

Keep smiling..!!

:)

Answer 2

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

$\green{\tt{\therefore{Area\:of\:triangle=6\:sq\:units}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{: \implies Coordinate \: of \: A= (2,0) } \\ \\ \tt{: \implies Coordinate \: of \: B = (-1,3) } \\ \\ \tt{: \implies Coordinate \: of \: C = (2,4) } \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies Area \: of \: triangle = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} | x_{1} ( y_{2} - y_{3}) + x_{2}( y_{3} - y_{1}) + x_{3}( y_{1} - y_{2} ) | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |2(3 - 4)-1(4 - 0) + 2(0 - 3)| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |2 \times -1 -1\times 4 + 2 \times -3 | } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} |-2-4 -6| } \\ \\ \tt{: \implies Area \: of \: triangle = \frac{1}{2} \times 12} \\ \\ \green{\tt{: \implies Area \: of \: triangle =6 \: sq \: units}} \\ \\ \purple{\bold{Some \: formula \: related \: to \: coordinate \: geometery}} \\ \pink{\tt{ \circ \: Distance \: formula = \sqrt{ (x_{2} - x_{1})^{2} + ( y_{2} - y_{1} )^{2} } }} \\ \\ \pink{\tt{ \circ \: Section \: formula = x= \frac{m x_{2} + n x_{1} }{m + n} }}$