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Ex 10.3

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

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

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Last updated at Feb. 3, 2020 by Teachoo

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)