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Question
Find the area of the triangle formed by the points
(–10, –4), (–8, –1) and (–3, –5)
Solution
Let the vertices be A(–10, –4), B(–8, –1) and C(–3, –5)
Area of ∆ABC = `1/2[(x_1y_2 + x_2y_3 + x_3y_1) - (x_2y_1 + x_3y_2 + x_1y_3)]`
= `1/2[(50 + 3 + 32) - (12 + 40 + 10)]`
= `1/2[((-8 xx -4) + (-10 xx -5) + (-3 xx -1)),(-(-1 xx -10) + (-4 xx -3) + (-5 xx -8))]`
= `1/2[85 - 62]`
= `1/2[23]`
= 11.5
Area of ∆ACB = 11.5 sq.units
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