Question

if the vertices of a triangle are (1, k), (4, -3) &(-9, 7) and it's area is 15 square units. find the value of K​

Answer 1

Answer :

• value of k is 3.

Solution :

Given :

• Vertices of the triangle are (1,k) , (4,-3) and (-9,7).
• Area of the triangle is 15unit².

To Find :

• Value of k.

$\\ \boxed{ \sf \: Area_{triangle} = \dfrac{1}{2} [ x_{1}(y_{2} -y _{3}) +x _{2}(y_{3} - y_{1}) + x_{3}(y_{1} -y _{2}) ]}$

We can find the value of k using the formula of Area of Triangle. We will substitute the given, the by simplifying we will get the value of k.

Here,

• (x₁,y₁) = (1,k)
• (x₂,y₂) = (4,-3)
• (x₃,y₃) = (-9,7)
• Area = 15 Unit.

$\\ \sf \: area = \dfrac{1}{2} [x _{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2}) ]$

$\LARGE\color{blue}\mathfrak{Substituting\: the\: values}$

$\\ \implies \sf \: 15 = \dfrac{1}{2} [1( - 3 - 7) + 4(7 - k) - 9(k + 3) ] \\ \\ \implies \sf \: 15 \times 2 = 1( - 10) + 4(7 - k) - 9(k + 3) \\ \\ \implies\sf \: 30 = - 10 + 28 - 4k - 9k - 27$

$\implies \sf \: 30 = - 9 - 13k \\ \\ \implies \sf \: 30 + 9 = 13k$

$\implies\sf \: 39 = 13k \\ \\ \implies \sf \: k = \cancel\frac{^3 39}{13} \\ \\\implies \underline{ \boxed{ \sf \: \pink{k = 3}}}$

Hence ,

value of k is 3.

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