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Ex 9.3

Ex 9.3, 1 (i)

Ex 9.3, 1 (ii) Important

Ex 9.3, 1 (iii)

Ex 9.3, 2 (i)

Ex 9.3, 2 (ii)

Ex 9.3, 2 (iii) Important

Ex 9.3, 3

Ex 9.3, 4 Important You are here

Ex 9.3, 5 (i) Important

Ex 9.3, 5 (ii)

Ex 9.3, 6

Ex 9.3, 7 Important

Ex 9.3, 8 Important

Ex 9.3, 9

Ex 9.3, 10

Ex 9.3, 11 Important

Ex 9.3, 12

Ex 9.3, 13 Important

Ex 9.3, 14

Ex 9.3, 15 Important

Ex 9.3, 16 Important

Ex 9.3, 17 Important

Question 1 (i) Deleted for CBSE Board 2024 Exams

Question 1 (ii) Deleted for CBSE Board 2024 Exams

Question 1 (iii) Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

Ex 9.3, 4 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)