1. Chapter 7 Class 10 Coordinate Geometry
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  3. Examples

Example 13 - Chapter 7 Class 10 Coordinate Geometry

Last updated at Dec. 8, 2016 by

Learn all Concepts of Chapter 7 Class 10 (with VIDEOS). Check - Coordinate Geometry - Class 10

Example 13 - Find area of triangle P(–1.5, 3), Q(6, -2) - Area of triangle

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

    3. Examples

    Ex 7.4 (Optional) →

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 [ 1.5( 6) + 6(1) + ( 3)(5) ] = 1/2 [ 9 + 6 15 ] = 1/2 [ 0 ] = 0

Chapter 7 Class 10 Coordinate Geometry
Serial order wise

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