


Ex 10.3
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 You are here
Ex 10.3, 6 (i) Important
Ex 10.3, 6 (ii)
Ex 10.3, 7
Ex 10.3, 8 Important
Ex 10.3, 9 Important
Ex 10.3, 10
Ex 10.3, 11
Ex 10.3, 12 Important
Ex 10.3, 13
Ex 10.3, 14 Important
Ex 10.3, 15
Ex 10.3, 16 Important
Ex 10.3, 17 Important
Ex 10.3, 18 Important
Ex 10.3
Ex 10.3, 5 Find the points on the x-axis, whose distances from the line π₯/3 + π¦/4 = 1 are 4 units. We need to find point on the x-axis Let any point on x-axis be P(x, 0) Given that perpendicular distance from point P(x, 0) from given line π₯/3 + π¦/4 = 1 is 4 Simplifying equation of line π₯/3 + π¦/4 = 1 (4π₯ + 3π¦ )/12 = 1 4x + 3y = 12 4x + 3y β 12 = 0 We know that Perpendicular distance from point (x, y) to the line Ax + By + C = 0 is d = |π΄π₯1 + π΅π¦1 + π|/β(π΄^2 + π΅^2 ) Given perpendicular distance of point P(x, 0) from line 4x + 3y β 12 = 0 is 4 Here x1 = x, y1 = 0 & A = 4 , B = 3 , C = β 12 & d = 4 Putting values 4 = |4(π₯) + 3(0) β 12|/β(γ(4)γ^2 + γ(3)γ^2 ) 4 = |4π₯ β 12|/β(16 + 9) 4 = |4π₯ β 12|/β25 4 = |4π₯ β 12|/5 4 Γ 5 = |4π₯β12| 20 = |4π₯β12| |4π₯β12| = 20 4x β 12 = Β± 20 Thus, 4x β 12 = 20 or 4x β 12 = β 20 4x β 12 = 20 4x = 20 + 12 4x = 32 x = 32/4 x = 8 4x β 12 = β20 4x = β20 + 12 4x = β8 x = (β8)/4 x = β2 4x β 12 = β20 4x = β20 + 12 4x = β8 x = (β8)/4 x = β2 Thus, x = 8 or x = β3 Hence the required points on x-axis are (8, 0) & (β2, 0)