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

Ex 9.3, 5 (i) Important You are here

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 2025 Exams

Question 1 (ii) Deleted for CBSE Board 2025 Exams

Question 1 (iii) Important Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

Ex 9.3, 6 Find the distance between parallel lines 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 We know that , distance between two parallel lines Ax + By + C1 = 0 & Ax + By + C2 = 0 is d = |𝐶_1− 𝐶_2 |/√(𝐴^2 + 𝐵^2 ) Equation of first line is 15x + 8y – 34 = 0 Above equation is of the form Ax + By + C1 = 0 where A = 15, B = 8 & C1 = − 34 Equation of second line is 15x + 8y + 31 = 0 Above equation is of the form Ax + By + C2 = 0 where A = 15 , B = 8 , C2 = 31 Distance between parallel lines 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 is d = |𝐶_1− 𝐶_2 |/√(𝐴^2 + 𝐵^2 ) Putting values d = |−34 − 31|/√(〖(15)〗^2 + (8)^2 ) d = |−34 − 31|/√(225 + 64) d = |−65|/√289 d = 65/√(17 × 17) d = 65/17 Thus, the required distance is 𝟔𝟓/𝟏𝟕 units