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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.