# Misc 5

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 5 Find the perpendicular distance from the origin to the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ) . Frist we find equation of line We know that equation of line joining two point (x1, y1) & (x2, y2) is (y – y1) = (𝑦2 − 𝑦1)/(𝑥2 − 𝑥1) (x – x1) Equation of line passing through (cos θ, sin θ) & (cos ϕ, sin ϕ) is (y – sin θ) = (sinϕ − sinθ)/(cosϕ − cosθ ) (x – cos θ) (cos ϕ – cos θ) (y – sin θ) = ("sin " ϕ" " − " sin " θ)(x − cos θ) cos ϕ (y – sin θ) − cos θ (y – sin θ) = "sin " ϕ (x − cos θ) − sin θ(x − cos θ) cos ϕy − cos ϕ sin θ − cos θy + cos θ sin θ = sin ϕ x − sin ϕ cos θ − sin θ x + sin θ cos θ cos ϕy − cos ϕ sin θ − cos θy + cos θ sin θ = sin ϕ x − sin ϕ cos θ − sin θ x + sin θ cos θ cos ϕ y − cos θ y − cos ϕ sin θ + cos θ sin θ = sin ϕ x − sin θ x − sin ϕ cos θ + sin θ cos θ (cos ϕ – cos θ)y − cos ϕ sin θ + cos θ sin θ = (sin ϕ – sin θ)x − sin ϕ cos θ + sin θ cos θ (cos ϕ – cos θ)y − (sin ϕ – sin θ)x = – "sin " ϕ" cos θ "+" sin θ cos θ "+ " cos " ϕ" sin θ" − "cos θ sin θ" (cos ϕ – cos θ)y − (sin ϕ – sin θ)x = − sin ϕ cos θ + cos ϕ sin θ − (sin ϕ – sin θ)x + (cos ϕ – cos θ)y = − sin ϕ cos θ + cos ϕ sin θ − (sin ϕ– sin θ)x + (cos ϕ – cos θ)y = cos ϕ sin θ − sin ϕ cos θ − (sin ϕ– sin θ)x + (cos ϕ - cos θ)y = sin θ cos ϕ − sin ϕ cos θ − (sin ϕ– sin θ)x + (cos ϕ - cos θ)y = sin (θ − ϕ) − (sin ϕ– sin θ)x + (cos ϕ - cos θ)y − sin (θ − ϕ) = 0 The above equation is of the form Ax + By + C = 0 Here A = − (sin ϕ – sin θ) , B = (cos ϕ – cos θ) , C = − sin (ϕ – θ) We know that distance of a point (x1, y1) from line Ax + By + C = 0 is d = |𝐴𝑥_1 + 𝐵𝑦_1 + 𝐶|/√(𝐴^2 + 𝐵^2 ) Distance of origin (0, 0) to the line − (sin ϕ– sin θ)x + (cos ϕ - cos 𝜃)y = sin (θ − ϕ) is d = |𝐴𝑥_1 + 𝐵𝑦_1 + 𝐶|/√(𝐴^2 + 𝐵^2 ) Putting values d = | − "(" sinϕ " –" sinθ ")" 0 + "(" cos ϕ" " − " " cos θ")" 0 + (−sin〖(θ − ϕ〗)|/√(〖"(" sinϕ " –" sinθ ")" 〗^2 + 〖"(" cos ϕ" " − " " cos θ")" 〗^2 ) d = |0 + 0 −〖 sin〗〖(θ − ϕ〗)|/√(〖"(" sinϕ " –" sinθ ")" 〗^2 + 〖"(" cos ϕ" " − " " cos θ")" 〗^2 ) d = |−〖 sin〗〖(θ − ϕ〗)|/√(〖"(" sinϕ " –" sinθ ")" 〗^2 + 〖"(" cos ϕ" " − " " cos θ")" 〗^2 ) d = |〖 sin〗〖(− θ + ϕ〗)|/√(〖"(" sinϕ " –" sinθ ")" 〗^2 + 〖"(" cos ϕ" " − " " cos θ")" 〗^2 ) d = |〖 sin〗〖(ϕ − θ〗)|/√(〖"(" sinϕ " –" sinθ ")" 〗^2 + 〖"(" cos ϕ" " − " " cos θ")" 〗^2 ) d = |〖 sin〗〖(ϕ − θ〗)|/√((2cos ((θ" " + " " ϕ)/2)". " sin ((ϕ" " − θ)/2))^2+(− 2 sin ((θ" " + " " ϕ)/2)"." sin((ϕ" " − θ)/2))^2 ) d = |〖 sin〗〖(ϕ − θ〗)|/√(4cos^2 ((θ" " + " " ϕ)/2)"." sin^2 ((ϕ" " − θ)/2)+ " " 〖4 sin〗^2 ((θ" " + " " ϕ)/2)"." sin^2 ((ϕ" " − θ)/2) ) d = |〖 sin〗〖(ϕ − θ〗)|/√(4〖 sin〗^2 ((ϕ" " − θ)/2)(cos^2 ((θ" " + " " ϕ)/2)+ 〖 sin〗^2 ((θ" " + " " ϕ)/2)) ) d = |〖 sin〗〖(ϕ − θ〗)|/(√(4〖 sin〗^2 ((ϕ" " − θ)/2).1) ) d = |〖 sin〗〖(ϕ − θ〗)|/√(2^2 〖 𝑠𝑖𝑛〗^2 ((ϕ" " − θ)/2) ) d = |〖 sin〗〖(ϕ − θ〗)|/(2 |𝑠𝑖𝑛((ϕ" " − θ)/2)| ) Thus, the required distance is |〖 sin〗〖(ϕ − θ〗)|/(2 |𝑠𝑖𝑛((ϕ" " − θ)/2)| )

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.