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Ex 10.3

Ex 10.3, 1 (i)

Ex 10.3, 1 (ii) Important

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Ex 10.3, 3 (i) Deleted for CBSE Board 2023 Exams

Ex 10.3, 3 (ii) Deleted for CBSE Board 2023 Exams

Ex 10.3, 3 (iii) Important Deleted for CBSE Board 2023 Exams

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Ex 10.3, 18 Important You are here

Last updated at Feb. 3, 2020 by Teachoo

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/π2 = 1/π2 + 1/π2 . Equation of line whose intercept on the axes are a & b is π₯/π + π¦/π = 1 The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by d = |π΄π₯_1 + π΅π¦_1 + πΆ|/β(π΄^2 + π΅^2 ) Now, π₯/π + π¦/π = 1 (1/π)x + (1/π)y β 1 = 0 Comparing with Ax + By + C = 0 Hence, A = 1/π, B = 1/π & C = β1 Also, Distance from origin (0, 0) to the line π₯/π + π¦/π = 1 is p So, distance d = p & x1 = 0, y1 = 0 Putting values d = |π΄π₯_1 + π΅π¦_1 + πΆ|/β(π΄^2 + π΅^2 ) p = |(0)(1/π) + (0)(1/π) β 1|/β((1/π)^2+ (1/π)^2 ) p = |0 + 0 β 1|/β(1/π2 + 1/π2) p = (|β1|)/β(1/π2 + 1/π2) p = 1/β(1/π2 + 1/π2) 1/π = β(1/π2+1/π2) Squaring both sides (1/π)^2 = (β(1/π2+1/π2))^2 π/ππ = π/ππ + π/ππ Hence proved