# Example 10 - Chapter 7 Class 10 Coordinate Geometry

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 10 If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p. Let the points be A(6, 1) , B(8, 2) C(9, 4) , D(p, 3) We know that diagonals of parallelogram bisect each other So, O is the mid−pint of AC & BD ∴ We find x co−ordinate of O from both AC & BD Finding mid−point of AC, We have to find x co−ordinate of O x−coordinate of O = (𝑥1 + 𝑥2)/2 Where x1 = 6 , x2 = 9 , Putting values for x−coordinate x−coordinate of O = (6 + 9)/2 = 15/2 Finding mid−point of BD, We have to find x co−ordinate of O x−coordinate of O = (𝑥1 + 𝑥2)/2 Where x1 = 8 , x2 = p , Putting values for x−coordinate x−coordinate of O = (8+ 𝑝)/2 Comparing (1) & (2) 15/2 = (8+ 𝑝)/2 15 = 8 + p 15 = 8 + p 15 – 8 = p 7 = p p = 7 Hence, p = 7

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

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.