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प्रश्न
Find the area of the triangle formed by the points
(1, – 1), (– 4, 6) and (– 3, – 5)
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उत्तर
Let the vertices A(1, – 1), B(– 4, 6) 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[((-4 xx - 5) + (-3 xx - 1) + (1 xx 6)),(-(-3 xx 6) + (1 xx -5) + (- 4 xx - 1))]`
= `1/2[(6 + 20 + 3) - (4 - 18 - 5)]`
= `1/2[29 - (- 19)]`
= `1/2[29 + 19]`
= `1/2 xx 48`
= 24 sq. units
Area of ∆ABC = 24 sq. units
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