Question

find the area of the traingle whose vertices(4,7) (1,3) (5,1)

Answer 1

4,7) = (x₁,y₁) ; (1,3) = (x₂,y₂) ; (5,1) = (x₃,y₃)

Area of the triangle, ∆ = \frac{1}{2}​2​​1​​ | x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂) |
= \frac{1}{2}​2​​1​​ | 4(3-1) + 1(1-7) + 5(7-3) |
= \frac{1}{2}​2​​1​​ | 4(2) + 1(-6) + 5(4) | 
= \frac{1}{2}​2​​1​​ | 8 - 6 + 20 |
= \frac{1}{2} | 2+20 |​2​​1​​∣2+20∣ 
= \frac{1}{2} | 22 |​2​​1​​∣22∣ 
= \frac{1}{2} (22)​2​​1​​(22) 
= 11

Therefore,the area of the triangle is 11 units².

Hope it helps