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Ex 7.3, 1 - Find area of triangle whose vertices are: - Area of triangle

  1. Chapter 7 Class 10 Coordinate Geometry
  2. Serial order wise
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Ex 7.3 , 1 Find the area of the triangle whose vertices are : (i) (2, 3), (–1, 0), (2, – 4) Let the two points be A(2, 3) , B(−1, 0) & C(2, −4) Area of triangle ABC = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = 2 , y1 = 3 x2 = −1 , y2 = 0 x3 = 2 , y3 = −4 Putting values Area of triangle ABC = 1/2 [ 2(0 – (−4)) + (−1)(−4 – 3 ) + 2(3 – 0) ] = 1/2 [ 2(4) + (−1)(– 7 ) + 2(3) ] = 1/2 [8 + 7 + 6] = 1/2 [21] = 10.5 square units 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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