Question

Find the area of the triangle whose vertices are (4 , 7) (1 , 3) (5 , 1)

Answer 1

area =1/2*[x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2)
=0.5*[4*(3-1)+1*(1-7)+5*(7-3)]
=0.5*(4*2+1*(-6)+5*4)
=0.5*(8-6+20)
=0.5*22
=11

Answer 2

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

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

Therefore,the area of the triangle is 11 units².

Hope it helps