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Ex 10.3

Ex 10.3, 1 (i)

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

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Ex 10.3, 12 Important

Ex 10.3, 13

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Ex 10.3, 16 Important

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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