1. Class 11
2. Important Question for exams Class 11

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 1/๐2 = 1/๐2 + 1/๐2 . Equation of line whose intercept on the axes are a & b is ๐ฅ/๐ + ๐ฆ/๐ = 1 The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by d = |๐ด๐ฅ_1 + ๐ต๐ฆ_1 + ๐ถ|/โ(๐ด^2 + ๐ต^2 ) Now, ๐ฅ/๐ + ๐ฆ/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 = โ1 Also, Distance from origin (0, 0) to the line ๐ฅ/๐ + ๐ฆ/๐ = 1 is p So, distance d = p & x1 = 0, y1 = 0 Putting values d = |๐ด๐ฅ_1 + ๐ต๐ฆ_1 + ๐ถ|/โ(๐ด^2 + ๐ต^2 ) p = |(0)(1/๐) + (0)(1/๐) โ 1|/โ((1/๐)^2 + (1/๐)^2 ) p = |0 + 0 โ 1|/โ(1/๐2 + 1/๐2) p = (| โ 1|)/โ(1/๐2 + 1/๐2) p = 1/โ(1/๐2 + 1/๐2) 1/๐ = โ(1/๐2 + 1/๐2) Squaring both sides (1/๐)^2 = (โ(1/๐2 + 1/๐2))^2 1/๐2 = 1/๐2 + 1/๐2 Hence proved

Class 11
Important Question for exams Class 11