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Question
Find the area of the triangle formed by the points
(–10, –4), (–8, –1) and (–3, –5)
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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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