Ex 9.3

Chapter 9 Class 11 Straight Lines
Serial order wise

### Transcript

Ex 9.3, 17 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/𝑝2 = 1/𝑎2 + 1/𝑏2 . Equation of line whose intercept on the axes are a & b is 𝑥/𝑎 + 𝑦/𝑏 = 1 The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by d = |𝐴𝑥_1 + 𝐵𝑦_1 + 𝐶|/√(𝐴^2 + 𝐵^2 ) Now, 𝑥/𝑎 + 𝑦/𝑏 = 1 (1/𝑎)x + (1/𝑏)y – 1 = 0 Comparing with Ax + By + C = 0 Hence, A = 1/𝑎, B = 1/𝑏 & C = –1 Also, Distance from origin (0, 0) to the line 𝑥/𝑎 + 𝑦/𝑏 = 1 is p So, distance d = p & x1 = 0, y1 = 0 Putting values d = |𝐴𝑥_1 + 𝐵𝑦_1 + 𝐶|/√(𝐴^2 + 𝐵^2 ) p = |(0)(1/𝑎) + (0)(1/𝑏) − 1|/√((1/𝑎)^2+ (1/𝑏)^2 ) p = |0 + 0 − 1|/√(1/𝑎2 + 1/𝑏2) p = (|−1|)/√(1/𝑎2 + 1/𝑏2) p = 1/√(1/𝑎2 + 1/𝑏2) 1/𝑝 = √(1/𝑎2+1/𝑏2) Squaring both sides (1/𝑝)^2 = (√(1/𝑎2+1/𝑏2))^2 𝟏/𝒑𝟐 = 𝟏/𝒂𝟐 + 𝟏/𝒃𝟐 Hence proved

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.