

Ā
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 9.3
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 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, 6 Find the distance between parallel lines 15x + 8y ā 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 = |š¶_1ā š¶_2 |/ā(š“^2 + šµ^2 ) Equation of first line is 15x + 8y ā 34 = 0 Above equation is of the form Ax + By + C1 = 0 where A = 15, B = 8 & C1 = ā 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 ā 34 = 0 and 15x + 8y + 31 = 0 is d = |š¶_1ā š¶_2 |/ā(š“^2 + šµ^2 ) Putting values d = |ā34 ā 31|/ā(ć(15)ć^2 + (8)^2 ) d = |ā34 ā 31|/ā(225 + 64) d = |ā65|/ā289 d = 65/ā(17 Ć 17) d = 65/17 Thus, the required distance is šš/šš units