Find the area of the triangle whose vertices are : ( I ) (2, 3), (–1, 0), (2, – 4) ( ii) (–5, –1), (3, –5), (5, 2).

(i) (2,3),(-1,0),(2,-4)
Vertices of triangle: A(2,3) ,B(-1,0) and C(2,-4)
Using the formula for area of triangle ∆ = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
Area of triangle ABC = 1/2[2{0-(-4)}+(-1){(-4)-3}+2(3-0)]
= 1/2[8+7+6]
= 21/2 = 10.5 square units
(ii) (-5,-1),(3,-5),(5,2)
Vertices of triangle: A(-5,-1), B(3,-5) and C(5,2)
Using the formula for area of trangle ∆ = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
Area of triangle ABC = 1/2 [(-5)(-5-2)+3{3(2-(-1)}+5{-1(-5)}]
= 1/2[35+9+20]
= 64/2 = 32 square units

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