Last updated at May 29, 2018 by Teachoo

Transcript

Ex10.3, 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/ 2 = 1/ 2 + 1/ 2 . Equation of line whose intercept on the axes are a & b is / + / = 1 The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by d = | _1 + _1 + |/ ( ^2 + ^2 ) Now, / + / = 1 (1/ )x + (1/ )y 1 = 0 Comparing with Ax + By + C = 0 Hence, A = 1/ , B = 1/ & C = 1 Also, Distance from origin (0, 0) to the line / + / = 1 is p So, distance d = p & x1 = 0, y1 = 0 Putting values d = | _1 + _1 + |/ ( ^2 + ^2 ) p = |(0)(1/ ) + (0)(1/ ) 1|/ ((1/ )^2 + (1/ )^2 ) p = |0 + 0 1|/ (1/ 2 + 1/ 2) p = (| 1|)/ (1/ 2 + 1/ 2) p = 1/ (1/ 2 + 1/ 2) 1/ = (1/ 2 + 1/ 2) Squaring both sides (1/ )^2 = ( (1/ 2 + 1/ 2))^2 1/ 2 = 1/ 2 + 1/ 2 Hence proved

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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.