# Misc 5 - Chapter 10 Class 11 Straight Lines (Term 1)

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 5 Find the perpendicular distance from the origin to the line joining the points (cos , sin ) and (cos , sin ) . Frist we find equation of line We know that equation of line joining two point (x1, y1) & (x2, y2) is (y y1) = ( 2 1)/( 2 1) (x x1) Equation of line passing through (cos , sin ) & (cos , sin ) is (y sin ) = (sin sin )/(cos cos ) (x cos ) (cos cos ) (y sin ) = ("sin " " " " sin " )(x cos ) cos (y sin ) cos (y sin ) = "sin " (x cos ) sin (x cos ) cos y cos sin cos y + cos sin = sin x sin cos sin x + sin cos cos y cos sin cos y + cos sin = sin x sin cos sin x + sin cos cos y cos y cos sin + cos sin = sin x sin x sin cos + sin cos (cos cos )y cos sin + cos sin = (sin sin )x sin cos + sin cos (cos cos )y (sin sin )x = "sin " " cos "+" sin cos "+ " cos " " sin " "cos sin " (cos cos )y (sin sin )x = sin cos + cos sin (sin sin )x + (cos cos )y = sin cos + cos sin (sin sin )x + (cos cos )y = cos sin sin cos (sin sin )x + (cos - cos )y = sin cos sin cos (sin sin )x + (cos - cos )y = sin ( ) (sin sin )x + (cos - cos )y sin ( ) = 0 The above equation is of the form Ax + By + C = 0 Here A = (sin sin ) , B = (cos cos ) , C = sin ( ) We know that distance of a point (x1, y1) from line Ax + By + C = 0 is d = | _1 + _1 + |/ ( ^2 + ^2 ) Distance of origin (0, 0) to the line (sin sin )x + (cos - cos )y = sin ( ) is d = | _1 + _1 + |/ ( ^2 + ^2 ) Putting values d = | "(" sin " " sin ")" 0 + "(" cos " " " " cos ")" 0 + ( sin ( )|/ ( "(" sin " " sin ")" ^2 + "(" cos " " " " cos ")" ^2 ) d = |0 + 0 sin ( )|/ ( "(" sin " " sin ")" ^2 + "(" cos " " " " cos ")" ^2 ) d = | sin ( )|/ ( "(" sin " " sin ")" ^2 + "(" cos " " " " cos ")" ^2 ) d = | sin ( + )|/ ( "(" sin " " sin ")" ^2 + "(" cos " " " " cos ")" ^2 ) d = | sin ( )|/ ( "(" sin " " sin ")" ^2 + "(" cos " " " " cos ")" ^2 ) d = | sin ( )|/ ((2cos (( " " + " " )/2)". " sin (( " " )/2))^2+( 2 sin (( " " + " " )/2)"." sin(( " " )/2))^2 ) d = | sin ( )|/ (4cos^2 (( " " + " " )/2)"." sin^2 (( " " )/2)+ " " 4 sin ^2 (( " " + " " )/2)"." sin^2 (( " " )/2) ) d = | sin ( )|/ (4 sin ^2 (( " " )/2)(cos^2 (( " " + " " )/2)+ sin ^2 (( " " + " " )/2)) ) d = | sin ( )|/( (4 sin ^2 (( " " )/2).1) ) d = | sin ( )|/ (2^2 ^2 (( " " )/2) ) d = | sin ( )|/(2 | (( " " )/2)| ) Thus, the required distance is | sin ( )|/(2 | (( " " )/2)| )

Miscellaneous

Misc 1
Important

Misc 2 Important

Misc 3 Important

Misc 4

Misc 5 You are here

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13

Misc 14

Misc 15 Important

Misc 16 Important

Misc 17

Misc 18 Important

Misc 19 Important

Misc 20 Important

Misc 21 Important

Misc 22 Important

Misc 23

Misc 24 Important

Chapter 10 Class 11 Straight Lines (Term 1)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.