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Ex 10.3, 5 - Find points on x-axis, whose distances from

Ex 10.3, 5 - Chapter 10 Class 11 Straight Lines - Part 2
Ex 10.3, 5 - Chapter 10 Class 11 Straight Lines - Part 3
Ex 10.3, 5 - Chapter 10 Class 11 Straight Lines - Part 4

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Ex 10.3, 5 Find the points on the x-axis, whose distances from the line π‘₯/3 + 𝑦/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 + 𝑦/4 = 1 is 4 Simplifying equation of line π‘₯/3 + 𝑦/4 = 1 (4π‘₯ + 3𝑦 )/12 = 1 4x + 3y = 12 4x + 3y – 12 = 0 We know that Perpendicular distance from point (x, y) to the line Ax + By + C = 0 is d = |𝐴π‘₯1 + 𝐡𝑦1 + 𝑐|/√(𝐴^2 + 𝐡^2 ) Given perpendicular distance of point P(x, 0) from line 4x + 3y – 12 = 0 is 4 Here x1 = x, y1 = 0 & A = 4 , B = 3 , C = βˆ’ 12 & d = 4 Putting values 4 = |4(π‘₯) + 3(0) βˆ’ 12|/√(γ€–(4)γ€—^2 + γ€–(3)γ€—^2 ) 4 = |4π‘₯ βˆ’ 12|/√(16 + 9) 4 = |4π‘₯ βˆ’ 12|/√25 4 = |4π‘₯ βˆ’ 12|/5 4 Γ— 5 = |4π‘₯βˆ’12| 20 = |4π‘₯βˆ’12| |4π‘₯βˆ’12| = 20 4x – 12 = Β± 20 Thus, 4x βˆ’ 12 = 20 or 4x βˆ’ 12 = βˆ’ 20 4x – 12 = 20 4x = 20 + 12 4x = 32 x = 32/4 x = 8 4x βˆ’ 12 = βˆ’20 4x = βˆ’20 + 12 4x = βˆ’8 x = (βˆ’8)/4 x = βˆ’2 4x βˆ’ 12 = βˆ’20 4x = βˆ’20 + 12 4x = βˆ’8 x = (βˆ’8)/4 x = βˆ’2 Thus, x = 8 or x = βˆ’3 Hence the required points on x-axis are (8, 0) & (βˆ’2, 0)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.