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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) Deleted for CBSE Board 2023 Exams
Ex 10.3, 3 (ii) Deleted for CBSE Board 2023 Exams
Ex 10.3, 3 (iii) Important Deleted for CBSE Board 2023 Exams
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
Last updated at March 16, 2023 by Teachoo
Ex 10.3, 5 Find the points on the x-axis, whose distances from the line 𝑥/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 𝑥/3 + 𝑦/4 = 1 is 4 Simplifying equation of line 𝑥/3 + 𝑦/4 = 1 (4𝑥 + 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 = |𝐴𝑥1 + 𝐵𝑦1 + 𝑐|/√(𝐴^2 + 𝐵^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 = − 12 & d = 4 Putting values 4 = |4(𝑥) + 3(0) − 12|/√(〖(4)〗^2 + 〖(3)〗^2 ) 4 = |4𝑥 − 12|/√(16 + 9) 4 = |4𝑥 − 12|/√25 4 = |4𝑥 − 12|/5 4 × 5 = |4𝑥−12| 20 = |4𝑥−12| |4𝑥−12| = 20 4x – 12 = ± 20 Thus, 4x − 12 = 20 or 4x − 12 = − 20 4x – 12 = 20 4x = 20 + 12 4x = 32 x = 32/4 x = 8 4x − 12 = −20 4x = −20 + 12 4x = −8 x = (−8)/4 x = −2 4x − 12 = −20 4x = −20 + 12 4x = −8 x = (−8)/4 x = −2 Thus, x = 8 or x = −3 Hence the required points on x-axis are (8, 0) & (−2, 0)