Slide52.JPG

Slide53.JPG

  1. Chapter 7 Class 10 Coordinate Geometry (Term 1)
  2. Serial order wise

Transcript

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 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.