# Ex 7.3, 1 (i) - 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

Ex 7.3 , 1 Find the area of the triangle whose vertices are : (i) (2, 3), (–1, 0), (2, – 4) Let the two points be A(2, 3) , B(−1, 0) & C(2, −4) Area of triangle ABC = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = 2 , y1 = 3 x2 = −1 , y2 = 0 x3 = 2 , y3 = −4 Putting values Area of triangle ABC = 1/2 [2(0 – (−4)) + (−1)(−4 – 3 ) + 2(3 – 0)] = 1/2 [ 2(4) + (−1)(– 7 ) + 2(3)] = 1/2 [8 + 7 + 6] = 1/2 [21] = 10.5 square units

Ex 7.3

Ex 7.3, 1 (i)
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Ex 7.3, 1 (ii) Important Deleted for CBSE Board 2022 Exams

Ex 7.3, 2 (i) Important Deleted for CBSE Board 2022 Exams

Ex 7.3, 2 (ii) Deleted for CBSE Board 2022 Exams

Ex 7.3, 3 Important Deleted for CBSE Board 2022 Exams

Ex 7.3, 4 Important Deleted for CBSE Board 2022 Exams

Ex 7.3, 5 Deleted for CBSE Board 2022 Exams

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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.