Question

If the vertices of the triangle are (1, k) (4, -3) (-9, 7) and its area is 15square unut then find the value of k.​

Answer 1

GIVEN :-

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

TO FIND :-

• Value of k.

TO KNOW :-

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

Here , (x₁,y₁) , (x₂,y₂) and (x₃,y₃) are co-ordinates of the vertices of triangle.

SOLUTION :-

Let,

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

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

We have , Area = 15unit².

Putting values,

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

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

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

Hence , value of k is 3.