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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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प्रश्न

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

योग

उत्तर

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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अध्याय 10: Straight Lines - Miscellaneous Exercise [पृष्ठ २३३]

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एनसीईआरटी Mathematics [English] Class 11
अध्याय 10 Straight Lines
Miscellaneous Exercise | Q 5 | पृष्ठ २३३

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