Check sibling questions

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


Transcript

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)

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.