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Ex 10.3, 18 - If p is length of perpendicular from origin

Ex 10.3, 18 - Chapter 10 Class 11 Straight Lines - Part 2
Ex 10.3, 18 - Chapter 10 Class 11 Straight Lines - Part 3 Ex 10.3, 18 - Chapter 10 Class 11 Straight Lines - Part 4

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Ex 10.3, 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/𝑝2 = 1/π‘Ž2 + 1/𝑏2 . Equation of line whose intercept on the axes are a & b is π‘₯/π‘Ž + 𝑦/𝑏 = 1 The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by d = |𝐴π‘₯_1 + 𝐡𝑦_1 + 𝐢|/√(𝐴^2 + 𝐡^2 ) Now, π‘₯/π‘Ž + 𝑦/𝑏 = 1 (1/π‘Ž)x + (1/𝑏)y – 1 = 0 Comparing with Ax + By + C = 0 Hence, A = 1/π‘Ž, B = 1/𝑏 & C = –1 Also, Distance from origin (0, 0) to the line π‘₯/π‘Ž + 𝑦/𝑏 = 1 is p So, distance d = p & x1 = 0, y1 = 0 Putting values d = |𝐴π‘₯_1 + 𝐡𝑦_1 + 𝐢|/√(𝐴^2 + 𝐡^2 ) p = |(0)(1/π‘Ž) + (0)(1/𝑏) βˆ’ 1|/√((1/π‘Ž)^2+ (1/𝑏)^2 ) p = |0 + 0 βˆ’ 1|/√(1/π‘Ž2 + 1/𝑏2) p = (|βˆ’1|)/√(1/π‘Ž2 + 1/𝑏2) p = 1/√(1/π‘Ž2 + 1/𝑏2) 1/𝑝 = √(1/π‘Ž2+1/𝑏2) Squaring both sides (1/𝑝)^2 = (√(1/π‘Ž2+1/𝑏2))^2 𝟏/π’‘πŸ = 𝟏/π’‚πŸ + 𝟏/π’ƒπŸ Hence proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.