Last updated at May 29, 2018 by Teachoo

Transcript

Ex10.3, 18 If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that 1/ 2 = 1/ 2 + 1/ 2 . Equation of line whose intercept on the axes are a & b is / + / = 1 The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by d = | _1 + _1 + |/ ( ^2 + ^2 ) Now, / + / = 1 (1/ )x + (1/ )y 1 = 0 Comparing with Ax + By + C = 0 Hence, A = 1/ , B = 1/ & C = 1 Also, Distance from origin (0, 0) to the line / + / = 1 is p So, distance d = p & x1 = 0, y1 = 0 Putting values d = | _1 + _1 + |/ ( ^2 + ^2 ) p = |(0)(1/ ) + (0)(1/ ) 1|/ ((1/ )^2 + (1/ )^2 ) p = |0 + 0 1|/ (1/ 2 + 1/ 2) p = (| 1|)/ (1/ 2 + 1/ 2) p = 1/ (1/ 2 + 1/ 2) 1/ = (1/ 2 + 1/ 2) Squaring both sides (1/ )^2 = ( (1/ 2 + 1/ 2))^2 1/ 2 = 1/ 2 + 1/ 2 Hence proved

Chapter 10 Class 11 Straight Lines

Ex 10.1, 5
Important

Ex 10.1, 7 Important

Ex 10.1, 9 Important

Ex 10.1, 13 Important

Ex 10.2, 8 Important

Ex 10.2, 14 Important

Ex 10.2, 18 Important

Example 15 Important

Ex 10.3, 5 Important

Ex 10.3, 8 Important

Ex 10.3, 10 Important

Ex 10.3, 16 Important

Ex 10.3, 18 Important You are here

Example 22 Important

Misc 6 Important

Misc 12 Important

Misc 18 Important

Misc 23 Important

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

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.