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Ex 10.3, 1 (ii) Important
Ex 10.3, 1 (iii)
Ex 10.3, 2 (i)
Ex 10.3, 2 (ii)
Ex 10.3, 2 (iii) Important
Ex 10.3, 3 (i)
Ex 10.3, 3 (ii)
Ex 10.3, 3 (iii) Important
Ex 10.3, 4
Ex 10.3, 5 Important
Ex 10.3, 6 (i) Important
Ex 10.3, 6 (ii)
Ex 10.3, 7
Ex 10.3, 8 Important
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Ex 10.3, 12 Important
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Ex 10.3, 14 Important
Ex 10.3, 15
Ex 10.3, 16 Important
Ex 10.3, 17 Important
Ex 10.3, 18 Important You are here
Ex 10.3
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