Last updated at May 29, 2018 by Teachoo

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Misc 12 Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes. Given lines are 4x + 7y − 3 = 0 2x – 3y + 1 = 0 We need to calculate Equation of line that passes through point of intersection of lines (1) & (2) & make equal intercepts on the axes Calculating point of intersection of lines (1) & (2) From (2) 2x − 3y + 1 = 0 2x = 3y − 1 x = (3𝑦 − 1)/2 Putting value of x in (1) 4x + 7y − 3 = 0 4((3𝑦 − 1)/2) + 7y − 3 = 0 2(3y − 1) + 7y − 3 = 0 6y − 2 + 7y − 3 = 0 13y − 5 = 0 13y = 5 y = 5/13 Putting value of y = 5/13 in (1) 2x − 3y + 1 = 0 2x − 3(5/13) + 1 = 0 2x − 15/13 + 1 = 0 2x = 15/13 – 1 2x = (15 − 13)/13 2x = 2/13 x = 2/(13 × 2) x = 1/13 Thus, point of intersection is (1/13, 5/13) Given that equation of line makes equal intercepts on the axes Let equation of line be 𝑥/𝑎 + 𝑦/𝑏 = 1 Where a is x − intercept & b is y − intercept Here both intercepts are same, So, b = a So equation of line becomes 𝑥/𝑎 + 𝑦/𝑎 = 1 x + y = a Also the line passes through (1/13, 5/13) Putting x = 1/13, y = 5/13 in our equation x + y = a 1/13 + 5/13 = a 6/13 = a a = 6/13 Thus, the equation of required lines becomes x + y = a x + y = 6/13 13(x + y) = 6 13x + 13y = 6 13x + 13y − 6 = 0

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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.