# Ex 10.3, 5 - Chapter 10 Class 11 Straight Lines

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 10.3, 5 Find the points on the x-axis, whose distances from the line ๐ฅ/3 + ๐ฆ/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 ๐ฅ/3 + ๐ฆ/4 = 1 is 4 Simplifying equation of line ๐ฅ/3 + ๐ฆ/4 = 1 (4๐ฅ + 3๐ฆ )/12 = 1 4x + 3y = 12 4x + 3y โ 12 = 0 We know that Perpendicular distance from point (x, y) to the line Ax + By + C = 0 is d = |๐ด๐ฅ1 + ๐ต๐ฆ1 + ๐|/โ(๐ด^2 + ๐ต^2 ) Given perpendicular distance of point P(x, 0) from line 4x + 3y โ 12 = 0 is 4 Here x1 = x , y1 = 0 & A = 4 , B = 3 , C = โ 12 & d = 4 Putting values 4 = |4(๐ฅ) + 3(0) โ 12|/โ(ใ(4)ใ^2 + ใ(3)ใ^2 ) 4 = |4๐ฅ โ 12|/โ(16 + 9) 4 = |4๐ฅ โ 12|/โ25 4 = |4๐ฅ โ 12|/5 4 ร 5 = |4๐ฅ โ 12| 20 = |4๐ฅ โ 12| |4๐ฅ โ 12| = 20 4x โ 12 = ยฑ 20 Thus, 4x โ 12 = 20 or 4x โ 12 = โ 20 Thus, x = 8 or x = โ 3 Hence the required points on x-axis are (8, 0) & ( โ 2, 0)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.