# Example 14

Last updated at Feb. 25, 2017 by Teachoo

Last updated at Feb. 25, 2017 by Teachoo

Transcript

Example 14 Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear. If the above points are collinear, they will lie on the same line, i.e. they will not form triangle Therefore, Area of ∆ABC = 0 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] = 0 Here x1 = 2 , y1 = 3 x2 = 4 , y2 = k x3 = 6 , y3 = −3 Putting values 1/2 [ 2(k – (−3)) + 4(−3 − 3) + 6(3 – k) ] = 0 2(k + 3) + 4(−6) + 6(3 − k) = 0 × 2 2k + 6 – 24 + 18 – 6k = 0 2k – 6k = − 6 + 24 −18 −4k = 0 ⇒ k = 0

Class 10

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.