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Question
Find the area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9).
Solution
Given that, the vertices of triangles are (–8, 4), (–6, 6) and (–3, 9).
Let (x1, y1) `→` (−8, 4)
(x2, y2) `→` (−6, 6)
And (x3, y3) `→` (−3, 9)
We know that, the area of triangle with vertices
(x1, y1), (x2, y2) and (x3, y3)
Δ = `1/2[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]`
= `1/2[-8(6 - 9) - 6(9 - 4) + (-3)(4 - 6)]`
= `1/2[-8(-3) - 6(5) - 3(-2)]`
= `1/2(24 - 30 + 6)`
= `1/2(30 - 30)`
= `1/2(0)`
= 0
Hence, the required area of triangle is 0.
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