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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
Ex 10.3, 6 (ii) You are here
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
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Ex 10.3, 6 Find the distance between parallel lines (ii) π(x + y) + p = 0 and π(x + y) β r = 0 We know that, distance between two parallel lines Ax + By + C1 = 0 & Ax + By + C2 = 0 is d = |πΆ_1 β πΆ_2 |/β(π΄^2 + π΅^2 ) Equation of the first line is π(x + y) + p = 0 πx + πy + p = 0 Above equation is of the form Ax + By + C1 = 0 where A = π , B = π & C1 = p Equation of the second line is π(x + y) β r = 0 πx + ly β r = 0 The above equation is of the form Ax + By + C2 = 0 where A = π , B = π , C2 = βr Distance between parallel lines π(x + y) + p = 0 & π(x + y) β r = 0 is d = |πΆ_1 β πΆ_2 |/β(π΄^2 + π΅^2 ) Putting values d = |π β (βπ)|/β(π^2 + π^2 ) d = |π + π|/β(2π^2 ) d = (|π + π| )/(|π|β2) d = |(π + π )/(πβ2)| Thus, the required distance is |(π + π )/(πβπ)| units