Last updated at May 29, 2018 by Teachoo

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Ex10.2, 18 P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is ๐ฅ/๐ + ๐ฆ/๐ = 2 Plotting x-axis and y-axis Let l be a line intersecting x-axis at A and y-axis at B Let P(a, b) be midpoint of AB Here Let co-ordinates of A be (p, 0) Let co-ordinates of B be (0, q) We know that mid point of a line joining points (x1, y1) & (x2, y2) is ((๐ฅ1 + ๐ฅ2 )/2, (๐ฆ1 + ๐ฆ2)/2) Mid point of a line joining points A (p,0) & B(0, q) is P(a,b) Putting values (a, b) = ((๐ + 0)/2, (0 + ๐)/2) (a, b) = (๐/2, ๐/2) So, p = 2a, q = 2b So, points A = (p, 0) = (2a, 0) B = (0, q) = (0, 2b) Finding equation of line by two point equation of line (y โ y1) = (๐ฆ_2 โ ใ ๐ฆใ_1)/(๐ฅ_2 โ ใ ๐ฅใ_1 ) (x โ x1) For equation of line l passing through (2a, 0) & (0, 2b) Here x1 = 2a, y1 = 0 x2 = 0, y2 = 2b Putting values (y โ y1) = (๐ฆ_2 โ ใ ๐ฆใ_1)/(๐ฅ_2 โ ใ ๐ฅใ_1 ) (x โ x1) (y โ 0) = (2๐ โ 0)/(0 โ 2๐) ( x โ 2a) y = 2๐/( โ 2๐) (x โ 2a) y = ( โ b)/a (x โ 2a) ay = โbx + 2ab ay + bx = 2ab Dividing by ab ๐๐ฆ/๐๐ + ๐๐ฅ/๐๐ = 2๐๐/2๐๐ ๐ฆ/๐ + ๐ฅ/๐ = 2 ๐ฅ/๐ + ๐ฆ/๐ = 2 Hence ,the equation of line is ๐ฅ/๐ + ๐ฆ/๐ = 2 Hence proved

Ex 10.1, 5
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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 You are here

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

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.