OX and OY are two coordinates axes. On OY is taken a fixed point P and on OX any point Q. On PQ an equilateral triangle is described, its vertex R being on the side of PQ away from O, then the locus of R will be -
Text Solution
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Let P be (0, c), c is a constant number
From the figure, h = OL = OQ + QL
= c cot θ + QR cos (180 – 60 – θ )
= c cot θ + PQ {cos 120 . cos θ + sin 120 . sin θ }
= c cot θ – 
cos θ + 
PQ sin θ
= c cot θ – 
c coses θ cos θ + 
c coses θ sin θ
= 
(cot θ + 
) … (i)
Again, k = RL = RQ sin (180 – 60 – θ ) = PQ
= c cosec θ 
= 
(1 + 
cot θ ) … (ii)
Eliminating cot θ from (i) and (ii), we get
k = 
h – c
∴ The required locus is y = 
x – c, a straight line.
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