(y – m 1 x) ( y – m 2 x) = c 1 c 2

Sol. Let the line (L) through the origin is

x = r cos θ ; y = r sin θ

as L intersects L 1 at Q and OQ = r 1

∴ r 1 sin θ = m 1 r 1 cos θ + c 1 ..............(1)

similarly, L intersects L 2 at R and OR = r 2

r 2 sin θ = m 2 r 2 cos θ + c 2 ..............(2)

Let P ≡ (h, k) & OP = r

∴ r 2 = r 1 r 2 ..............(3)

& h = r cos θ ..............(4)

k = r sin θ ..............(5)

putting the values of r 1 and r 2 from (1) and (2) in (3)

∴ r 2 = . ..............(6)

putting the value of cos θ and sin θ from (4) and (5) in (6), we get

⇒ r 2 = ⇒ (k – m 1 h) (k – m 2 h) = c 1 c 2

replacing (h, k) by (x, y) we get the desired locus as (y – m 1 x) ( y – m 2 x) = c 1 c 2

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