NCERT Solutions
Class 11 Maths
Straight Lines
Ex.Misc. Q.5
Find the perpendicular distance from the origin to the line joining the points (cos θ, sin θ) and (cos ф, sin ф)
The equation of the line joining the points (cos θ, sin θ) and (cos ф, sin ф) is given by
y – sin θ = {(sin ф - sin θ) ÷ (cos ф - cos θ)} (x – sin θ)
=> (y – sin θ) (cos ф - cos θ) = (sin ф - sin θ) (x – cos θ)
=> (cos ф - cos θ) y – sin θ (cos ф - cos θ) = x (sin ф - sin θ) – cos θ (sin ф - sin θ)
=> x (sin θ - sin ф) + y (cos ф - cos θ) + cos θ sin ф – cos θ sin θ –sin θ cos ф + sin θ cos θ = 0
=> x (sin θ - sin ф) + y (cos ф - cos θ) + sin (θ – ф) = 0
=> Ax + By + C = 0
Where A = sin θ - sin ф, B = cos ф - cos θ and C = sin (θ – ф)
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is
given by d = |Ax1 + By1 + C| ÷ √ (A2 + B2)
Therefore, the perpendicular distance (d) of the given line from point (x1, y1) = (0, 0) is
d = [| (sin θ - sin ф) × 0 + (cos ф - cos θ) × 0 + sin (θ – ф) |] ÷ [√ {(sin θ - sin ф)2 + (cos ф - cos θ)2}]
=> d = |sin (θ – ф) | ÷ √ {sin2 θ + sin2 ф - 2 sin θ × sin ф + cos2 ф + cos2 θ - 2 cos ф × cos θ}
=> d = |sin (θ – ф) |÷ √ {sin2 θ + cos2 θ - 2 sin θ × sin ф + sin2 ф + cos2 ф - 2 cos ф × cos θ}
=> d = |sin (θ – ф) | ÷ √ (1 - 2 sin θ × sin ф + 1 - 2 cos ф × cos θ)
=> d = |sin (θ – ф) |÷ √ (2 - 2 sin θ × sin ф - 2 cos ф × cos θ)
=> d = |sin (θ – ф) | ÷ √ {2(1 - cos (ф - θ)}
=> d = |sin (θ – ф) | ÷ √ {2(2sin2 (ф - θ) ÷2}
=> d = |sin (θ – ф) | ÷ |2sin (ф - θ) ÷ 2|