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# Find the perpendicular distance from the origin to the line joining the points (cosθ, sinθ) and (cosΦ, sinΦ)

**Solution:**

The equation of the line joining the points (cosθ, sinθ) and (cosΦ, sinΦ) is given by

(y - sinθ)/(x - cosθ) = (sinΦ- sinθ)/(cosΦ - cosθ)

y - sinθ = (sinΦ - sinθ)/(cosΦ - cosθ) × (x - cosθ)

y (cosΦ - cosθ) - sinθ (cosΦ - cosθ) = x (sinΦ - sinθ) - cosθ (sinΦ - sinθ)

-x (sinΦ - sinθ) + y (cosΦ - cosθ) + cosθ sinΦ - cosθ sinθ - sinθ cosΦ + sinθ cosθ = 0

-x (sinΦ - sinθ) + y (cosΦ - cosθ) + cosθ sinΦ - sinθ cosΦ = 0

-x (sinΦ - sinθ) + y (cosΦ - cosθ) + sin (Φ - θ) = 0

Ax + By + C = 0 , where A = sinθ - sinΦ, B = cosΦ - cosθ and C = sin(Φ - θ)

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x\(_1\), y\(_1\)) is given by

d = |Ax\(_1\) + By\(_1\) + C|/√A² + B²

Therefore, the perpendicular distance (d) of the given line from point (x_{1}, y_{1}) = (0, 0) is

d = |(sinΦ - sinθ)(0) + (cosΦ - cosθ)(0) + sin(Φ - θ)|/√(sinθ - sinΦ)² + (cosΦ - cosθ)²

= |sin(Φ - θ)|/√(sin² Φ + sin² θ - 2 sinθ sin Φ + cos² Φ + cos² θ - 2 cosΦ cosθ)

= |sin(Φ - θ)|/√(sin² Φ + cos² Φ) + (sin² θ + cos² θ) - 2 (sinθ sin Φ + cosΦ cosθ)

= |sin(Φ - θ)|/√1 + 1 - 2 (cos (Φ - θ))

= |sin(Φ - θ)|/√2 (1 - cos (Φ - θ))

= |sin(Φ - θ)|/√2 (2 sin² (Φ - θ)/2)

= |sin(Φ - θ)|/|2 sin (Φ - θ)/2|

We know that sin A = 2 (cos A/2 sin A/2)

Then,

= |2 (cos(Φ - θ)/2 sin(Φ - θ)/2)|/|2 sin (Φ - θ)/2|

= |cos(Φ - θ)/2|

NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 5

## Find the perpendicular distance from the origin to the line joining the points (cosθ, sinθ) and (cosΦ, sinΦ)

**Summary:**

The perpendicular distance from the origin to the line joining the points (cosθ, sinθ) and (cosΦ, sinΦ) is |cos(Φ - θ)/2|

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