Misc 19 - Chapter 10 Class 11 Straight Lines

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Misc 19 If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m. Let line AB be y = 3x + 1 ,line CD be 2y = x + 3 & line PQ be y = mx + 4 Lines AB & CD are equally inclined to the line PQ First we find slopes of lines Slope of line AB y = 3x + 1 The above equation is of the form y = m1x + c1 where slope of line AB = m1 Slope of line AB = m1 = 3 Slope of line CD 2y = x + 3 y = ( + 3)/2 y = (1/2)x + 3/2 The above equation of the form y = m2x + c2 where slope of line CD = m2 Slope of CD = m2 = 1/2 Slope of line PQ y = mx + 4 The above equation of the form y = mx + c where m is slope Thus, Slope of line PQ = m Now, given that lines AB & CD are equally inclined to the line PQ i.e. line AB & line CD make equal angle with line PQ Angle between AB & PQ = angle between CD & PQ We know that angle between two lines whose slope are m1 & m2 is tan = |( _2 _1)/(1 + _2 _1 )| Since angles ( ) are equal So, tan must be equal From (A) & (B) |( 3)/(1 + 3 )| = |(2 1)/(2 + )| So, (( 3)/(1 + 3 )) = ((2 1)/(2 + )) or (( 3)/(1 + 3 )) = ((2 1)/(2 + )) Solving (( )/( + ))= (( )/( + )) ( 3)/(1 + 3 ) = (2 1)/(2 + ) (m 3)(2 + m) = (2m 1) (1 + 3m) m (2 + m) 3(2 + m) = 2m + 6m2 1 + 3m 5m + m2 6 = 5m + 6m2 1 5m 5m + m2 6m2 6 + 1 = 0 0 5m2 5 = 0 5m2 = 5 m2 = 5/( 5) m2 = 1 This is not possible as square of a number cannot be negative Solving ( )/( + ) = (( )/( + )) ( 3)/(1 + 3 ) = (1 2 )/(2 + ) (m 3) (2 + m) = (1 2m)(1 + 3m) m(2 + m) 3(2 + m) = 1(1 + 3m) 2m(1 + 3m) 2m + m2 6 3m = 1 + 3m 2m 6m2 m2 6 m = 6m2 + m + 1 m2 + 6m2 m m 6 1 = 0 7m2 2m 7 = 0 The above equation is of the form ax2 + 3x + c = 0 Where solution is x = ( ( ^2 4 ))/2 Here a = 7, b = 2,c = 7 & x = m So, m = ( ( 2) ( ( 2) ^2 4(7)( 7)))/(2(7)) m = (2 (4 (1 + (7) ( 7)))/14 m = (2 2 ((1 + 49)))/14 m = (2(1 50 ))/14 m = (2(1 (5 5 2)))/14 m = 1/7(1 5 2) m = (1 5 2)/7 Thus, the required value of m is (1 + 5 2)/7 & (1 5 2)/7