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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)
Ex 10.3, 7
Ex 10.3, 8 Important
Ex 10.3, 9 Important You are here
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 Sept. 6, 2021 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Ex10.3, 9 Find angles between the lines 3x + y = 1 and x + 3 y = 1 Given equation of lines, 3x + y = 1 x + 3 y = 1 We know that angle between 2 lines ( ) can be found by using formula tan =|( _2 _1)/(1 + _2 _1 )| Let slope of line (1) be m1 & slope of line (2) be m2 Now, Angle between lines 3x + y = 1 and x + 3y = 1 is tan =|( _2 _1)/(1 + _2 _1 )| Putting values tan = |( 3 (( 1)/ 3))/(1 + ( 3)(( 1)/ 3) )| tan = |( 3 + 1/ 3)/(1 + 1)| tan = |((( 3)( 3) + 1)/ 3)/2| tan = |( 3 + 1)/(2 3)| tan = |( 2)/(2 3)| tan = |( 1)/ 3| tan = 1/ 3 We know that tan 30 = 1/ 3 tan = tan 30 Hence = 30 Thus, the acute angle between the lines (1) & (2) is = 30 & obtuse angle ( ) between these two lines is = 180 = 180 30 = 150 Thus, the required angle between lines is 30 or 150