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Example 13 Find the area of the triangle formed by the points P(–1.5, 3), Q(6, –2) and R(–3, 4). Area of triangle PQR = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = −1.5 , y1 = 3 x2 = 6 , y2 = −2 x3 = −3 , y3 = 4 Putting values Area of triangle PQR = 1/2 [ (−1.5)(−2 – 4) + 6(4 – 3 ) + (−3)(3 – (−2)) ] = 1/2 [ −1.5(−6) + 6(1) + (−3)(3 + 2) ] = 1/2 [ −1.5(−6) + 6(1) + (−3)(5) ] = 1/2 [ 9 + 6 − 15 ] = 1/2 [ 0 ] = 0 square units Note: Since Area of Triangle is 0 square units It means that the three vertices are collinear

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.