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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 7.3 , 1 Find the area of the triangle whose vertices are : (ii) (–5, –1), (3, –5), (5, 2) Let the two points be A(−5, −1) , B(3, −5) & C(5, 2) Area of triangle ABC = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] Here x1 = −5 , y1 = −1 x2 = 3 , y2 = −5 x3 = 5 , y3 = 2 Putting values Area of triangle ABC = 1/2 [ (−5)(−5 − 2) + 3(2 – (−1)) + 5(−1 – (−5))] = 1/2 [ −5(−7) + 3(2 + 1 ) + 5(−1 + 5)] = 1/2 [ −5(−7) + 3(3) + 5(4) ] = 1/2 [ 35 + 9 + 20] = 1/2 [64] = 32 square units

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.