Question

Find the value of k for which the area formed by the triangle with vertices ( 1, 2), ( -2, 3) and ( -3, k) is 11 square units

Answer 1

Given:-

๛A=(1,2)

๛B=(-2,3)

๛C=(-3,k)

๛Also, Area of ∆ˡᵉ=11sq.units

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To Find:-

→Find the Value of k?

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Answer:-

•Using the Formula•

→½×[x¹(y²-y³)+x²(y³-y¹)+x³(y¹-y²)]

★Putting the Values★

→½×[1(3-k)+-2(k-2)+-3(2-3)]=11

→½×[3-k-2k+2+3]=11

→-3k+2=11×2

→-3k=22-2

→-3k=20

→k=$\frac{\cancel{20}}{\cancel{-3}}$

→k=-6.66[Approx]

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Answer 2

$\tt{\huge{\pink{Hello!!! }}}$

$\tt{\huge{\boxed{\boxed{\orange{\bullet{Question \: - }}}}}}$

Find the value of k for which the area formed by the triangle with vertices ( 1, 2), ( -2, 3) and ( -3, k) is 11 square units.

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$\tt{\blue{K \: = \dfrac{-1}{3} }} .......(A_{1})$

$\tt{\red{K \: = 7 }} .......(A_{2})$

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$\tt{\purple {Step-By-Step-Explaination \: :- }}$

The solution is in the attachment.....

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