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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10

Find the area of the triangle formed by the points (1, – 1), (– 4, 6) and (– 3, – 5) - Mathematics

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Question

Find the area of the triangle formed by the points

(1, – 1), (– 4, 6) and (– 3, – 5)

Sum

Solution

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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Chapter 5: Coordinate Geometry - Exercise 5.1 [Page 211]

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Samacheer Kalvi Mathematics [English] Class 10 SSLC TN Board
Chapter 5 Coordinate Geometry
Exercise 5.1 | Q 1. (i) | Page 211
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