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Find perpendicular distance from the origin to the line joining the points (cosΘ, sin Θ) and (cosΦ, sin Φ). - Mathematics

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Question

Find perpendicular distance from the origin to the line joining the points (cosΘ, sin Θ) and (cosΦ, sin Φ).

Sum
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Solution

The equation of the line joining the points (cosθ, sinθ) and (cos∅, sin∅) is given by

= `y - sin θ = (sin∅ - sinθ)/(cos∅ - cosθ) (x - cosθ)`

= y(cos∅ - cosθ)-sinθ(cos∅ - cosθ) = x(sin∅ - sinθ)-cosθ (sin∅ - sinθ)

= x(sinθ - sin∅)+y(cos∅ - cosθ) + cosθ sin∅ - cosθ sinθ - sinθ cos∅ + sinθ cosθ  = 0

= x(sinθ - sin∅)+y(cos∅ - cosθ) + sin (∅ - θ) = 0

= Ax + By + C = 0, where A = sin θ - sin∅, B = cos∅ - cosθ, and C = sin (∅ - θ)

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

`d = |Ax_1 + By_1 + C|/sqrt(A^2 + B^2)`

Therefore, perpendicular distance (d) of the given line from the point (x1, y1) = (0, 0) is 

`d = |(sinθ - sin∅)(0) + (cos∅ - cosθ)(0) + sin(∅ - θ)|/sqrt((sinθ - sin∅)^2 + (cos∅ - cosθ)^2`

= `|sin (∅ - θ)|/sqrt (sin^2θ + sin^2∅ - 2sinθ sin∅ + cos^2∅ + cos^2θ - 2cos∅ cosθ)`

=  `|sin (∅ - θ)|/sqrt ((sin^2θ + cos^2θ) - (sin^2∅ cos^2∅) -2(sinθ - sin∅ + cosθ cos∅)`

= `|sin (∅ - θ)|/sqrt(1 + 1 - 2(cos (∅ - θ)))`

= `|sin (∅ - θ)|/sqrt(2(1 - cos (∅ - θ))`

= `|sin (∅ - θ)|/sqrt(2(2sin^2 ((∅ - θ)/2))`

= `|sin (∅ - θ)|/(|2sin((∅ - θ)/2)|)`

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Chapter 10: Straight Lines - Miscellaneous Exercise [Page 233]

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NCERT Mathematics [English] Class 11
Chapter 10 Straight Lines
Miscellaneous Exercise | Q 5 | Page 233

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