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Ex 9.3

Ex 9.3, 1 (i)

Ex 9.3, 1 (ii) Important

Ex 9.3, 1 (iii)

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Ex 9.3, 2 (iii) Important

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Ex 9.3, 8 Important You are here

Ex 9.3, 9

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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

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