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