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Example 13 - Chapter 7 Class 10 Coordinate Geometry (Term 1)
Last updated at Aug. 16, 2021 by Teachoo
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Chapter 7 Class 10 Coordinate Geometry
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
