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Question 2 - Area of triangle - Chapter 7 Class 10 Coordinate Geometry
Last updated at April 16, 2024 by Teachoo
Chapter 7 Class 10 Coordinate Geometry
Concept wise
- Distance Formula
- Equidistant points
- Checking points collinear or not
- Type of triangle formed
- Type of quadrilateral formed
- Section Formula- Finding coordinates
- Section Formula- Finding coordinates of a point in a quadrilateral
- Finding ratio
-
Area of triangle
- Given area, finding k
- Area of quadrilateral
Transcript
Question 2 Find the area of a triangle formed by the points A(5, 2), B(4, 7) and C (7, – 4). Area of triangle ABC = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = 5 , y1 = 2 x2 = 4 , y2 = 7 x3 = 7 , y3 = −4 Putting values Area of triangle ABC = 1/2 [ 5(7 – (−4)) + 4(−4 – 2 ) + 7(2 − 7) ] = 1/2 [ 5(7 + 4) + 4(−6 ) + 7(−5) ] = 1/2 [ 5(11) + 4(−6 ) + 7(−5) ] = 1/2 [55 – 24 – 35 ] = 1/2 [−4] = −2 Since, Area cannot be negative Area of triangle ABC = 2 square units
