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Last updated at Feb. 3, 2020 by Teachoo

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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 1/๐2 = 1/๐2 + 1/๐2 . Equation of line whose intercept on the axes are a & b is ๐ฅ/๐ + ๐ฆ/๐ = 1 The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by d = |๐ด๐ฅ_1 + ๐ต๐ฆ_1 + ๐ถ|/โ(๐ด^2 + ๐ต^2 ) Now, ๐ฅ/๐ + ๐ฆ/๐ = 1 (1/๐)x + (1/๐)y โ 1 = 0 Comparing with Ax + By + C = 0 Hence, A = 1/๐, B = 1/๐ & C = โ1 Also, Distance from origin (0, 0) to the line ๐ฅ/๐ + ๐ฆ/๐ = 1 is p So, distance d = p & x1 = 0, y1 = 0 Putting values d = |๐ด๐ฅ_1 + ๐ต๐ฆ_1 + ๐ถ|/โ(๐ด^2 + ๐ต^2 ) p = |(0)(1/๐) + (0)(1/๐) โ 1|/โ((1/๐)^2+ (1/๐)^2 ) p = |0 + 0 โ 1|/โ(1/๐2 + 1/๐2) p = (|โ1|)/โ(1/๐2 + 1/๐2) p = 1/โ(1/๐2 + 1/๐2) 1/๐ = โ(1/๐2+1/๐2) Squaring both sides (1/๐)^2 = (โ(1/๐2+1/๐2))^2 ๐/๐๐ = ๐/๐๐ + ๐/๐๐ Hence proved

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.