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Find the area of triangle whose vertices are A(5, 8), B(5, 0) C(1, 0)

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Question

Find the area of triangle whose vertices are

A(5, 8), B(5, 0) C(1, 0)

Sum
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Solution

Here, A(x1, y1) ≡ A(5, 8), B(x2, y2) ≡ B(5, 0), C(x3, y3) ≡ C(1, 0)

Area of a triangle = `1/2|(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|`

∴ A(ΔABC) = `1/2|(5, 8, 1),(5, 0, 1),(1, 0, 1)|`

= `1/2[5(0 - 0) - 8(5 - 1) + 1(0 - 0)]`

= `1/2[0 - 8(4) + 0]`

= `1/2(-32)`

= – 16

Since area cannot be negative,
A(ΔABC) = 16 sq. units

shaalaa.com
Application of Determinants - Area of Triangle and Collinearity of Three Points
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Chapter 4: Determinants and Matrices - Exercise 4.3 [Page 75]

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