# Example 11 - Chapter 7 Class 10 Coordinate Geometry (Term 1)

Last updated at Aug. 16, 2021 by Teachoo

Last updated at Aug. 16, 2021 by Teachoo

Transcript

Example 11 Find the area of a triangle whose vertices are (1, –1), (– 4, 6) and (–3, –5). Let the two points be A (1, −1) , B(−4, 6) & C(−3, −5) Area of triangle ABC = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = 1 , y1 = −1 x2 = −4 , y2 = 6 x3 = −3 , y3 = −5 Putting values Area of triangle ABC = 1/2 [ 1(6 – (−5)) + (−4)(−5 – (−1) ) + (−3)(−1 − 6) ] Area of triangle ABC = 1/2 [1(6 – (−5)) + (−4)(−5 – (−1) ) + (−3)(−1 − 6)] = 1/2 [1(6 + 5) + (−4)(−5 + 1 ) + (−3)(−7)] = 1/2 [ 1(11) + (−4)(−4 ) + (−3)(−7) ] = 1/2 [ 11 + 16 + 21 ] = 1/2 [ 48 ] = 24 square units

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Example 2 Important

Example 3 Important

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Example 7 Important

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Example 9 Important

Example 10 Important

Example 11 Deleted for CBSE Board 2022 Exams You are here

Example 12 Deleted for CBSE Board 2022 Exams

Example 13 Important Deleted for CBSE Board 2022 Exams

Example 14 Important Deleted for CBSE Board 2022 Exams

Example 15 Deleted for CBSE Board 2022 Exams

Chapter 7 Class 10 Coordinate Geometry (Term 1)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.