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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
Ex 10.3, 6 (i) Important You are here
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 Aug. 28, 2021 by Teachoo
Ā
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Ex 10.3, 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