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Example 13 Important Deleted for CBSE Board 2022 Exams You are here
Example 14 Important Deleted for CBSE Board 2022 Exams
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Last updated at Aug. 16, 2021 by Teachoo
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