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Misc 5 - Find perpendicular distance from origin to line - Miscellaneous

Misc 5 - Chapter 10 Class 11 Straight Lines - Part 2
Misc 5 - Chapter 10 Class 11 Straight Lines - Part 3 Misc 5 - Chapter 10 Class 11 Straight Lines - Part 4 Misc 5 - Chapter 10 Class 11 Straight Lines - Part 5 Misc 5 - Chapter 10 Class 11 Straight Lines - Part 6


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)| )

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.