NCERT Solutions
Class 11 Maths
Straight Lines
Ex.10.3 Q.16
If p and q are the lengths of perpendiculars from the origin to the lines
x cos θ – y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively,
prove that p2 + 4q2 = k2
The equations of given lines are
x cos θ – y sinθ = k cos 2θ ................... (1)
x sec θ + y cosec θ = k ......................... (2)
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)
On comparing equation 1 to the general equation of line i.e., Ax + By + C = 0, we get
A = cos θ, B = –sin θ, and C = –k cos 2θ
It is given that p is the length of the perpendicular from (0, 0) to line 1.
So, p = |A × 0 + B × 0 + C| ÷ √ (A2 + B2)
= |C| ÷ √ (cos2 θ + sin2 θ)
= |-k cos 2θ| ÷ √ (cos2 θ + sin2 θ)
= |-k cos 2θ|
It is given that q is the length of the perpendicular from (0, 0) to line 2.
So, q = |A × 0 + B × 0 + C| ÷ √ (A2 + B2)
= |C| ÷ √ (sec2 θ + cosec2 θ)
= |-k| ÷ √ (sec2 θ + cosec2 θ)
From equation (3) and (4), we have
p2 + 4q2 = (|-k cos 2θ|)2 + 4[|-k| ÷ √ (sec2 θ + cosec2 θ)]2
= k2 cos2 2θ + 4 k2 ÷ (sec2 θ + cosec2 θ)
= k2 cos2 2θ + 4 k2 ÷ ((1 ÷ cos2 θ) + (1 ÷ sin2 θ))
= k2 cos2 2θ + 4 k2 ÷ {(cos2 θ + sin2 θ) ÷ (cos2 θ × sin2 θ)}
= k2 cos2 2θ + 4 k2 ÷ {1 ÷ (cos2 θ × sin2 θ)}
= k2 cos2 2θ + 4 k2 (cos2 θ × sin2 θ)
= k2 cos2 2θ + k2 (4cos2 θ × sin2 θ)
= k2 cos2 2θ + k2 (2cos θ × sin θ) 2
= k2 cos2 2θ + k 2 sin 2 2θ
= k2(cos2 2θ + sin 2 2θ)
= k2
Hence, we proved that p2 + 4q2 = k2