Question

Points E (1, 3), F (9, 4) and G (2, 9) form a triangle. What is the area of ΔEFG ?
The area of ΔEFG = ____ square units

Please answer this question

Answer 1

Answer:-

Given:

E(1 , 3) , F(9 , 4) , G(2 , 9) form a triangle.

We know that,

Area of a ∆ formed with the points (x₁ , y₁) , (x₂ , y₂) & (x₃ , y₃) is :

$\implies \boxed{\sf \: Area \: of \: the \: \triangle = \dfrac{1}{2} \begin{vmatrix} \sf \: x _{1} - x _{2}& \sf \: x _{1} - x _{3} \\ \\ \sf \: y _{1} - y _{2}& \sf \: y _{1} - y _{3} \end{vmatrix}}$

Let,

• x₁ = 1

• x₂ = 9

• x₃ = 2

• y₁ = 3

• y₂ = 4

• y₃ = 9

$\implies{\sf \: Area \: of \: \triangle EFG \: = \dfrac{1}{2} \begin{vmatrix} \sf \: 1 - 9& \sf \: 1 - 2\\ \\ \sf \: 3 - 4& \sf \: 3 - 9 \end{vmatrix}} \\ \\ \implies{\sf \: Area \: of \: \triangle EFG \: = \dfrac{1}{2} \begin{vmatrix} \sf \: - 8& \sf \: - 1 \\ \\ \sf \: - 1& \sf \: - 6 \end{vmatrix}} \\ \\ \implies\sf \: Area \: of \: \triangle EFG \: = \dfrac{1}{2} |( - 8)( - 6) - ( - 1)( - 1)| \\ \\ \implies\sf \: Area \: of \: \triangle EFG \: = \dfrac{1}{2} |48 - 1| \\ \\ \implies \boxed{\sf \: Area \: of \: \triangle EFG \: = \dfrac{47}{2} \: \: unit ^{2} }$