Maharashtra State BoardHSC Commerce 11th
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Find the area of triangles whose vertices are A(− 1, 2), B(2, 4), C(0, 0) - Mathematics and Statistics

Sum

Find the area of triangles whose vertices are A(− 1, 2), B(2, 4), C(0, 0)

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Solution

Here, A(x1, y1) ≡ A(–1, 2), B(x2, y2) ≡ B(2, 4), C(x3, y3) ≡ C(0, 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|(-1, 2, 1),(2, 4, 1),(0, 0, 1)|`

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

= `1/2(-4 - 4)`

= `1/2(- 8)`

= – 4
Since, area cannot be negative.
∴ A(ΔABC) = 4 sq. units

Concept: Application of Determinants - Area of a Triangle
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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 11th Standard Maharashtra State Board
Chapter 6 Determinants
Miscellaneous Exercise 6 | Q 7. (i) | Page 95
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