`O X` and `O Y` are two coordinate axes. On `O Y` is taken a fixed point `P(0,c)` and on `O X` any point `Qdot` On `P Q ,` an equilateral triangle is described, its vertex `R` being on the side of `P Q` away from `O` . Then prove that the locus of `R` is `y=sqrt(3)x-cdot`

Explanation

This given fixed point is `P(0,c)`.
From given fig.,

`h=OL=OQ+QL`
`=c cot theta+Qrcos(180^@-60^@-theta)`
`=c cot theta+ PQ {cos120^@.costheta+sin120^@.sintheta}`
`=c cot theta-(PQ)/(2) costheta + (sqrt3)/(2)PQ sintheta`
`=ccot theta -(1)/(2) c cosec theta cos theta +(sqrt3)/(2) c cosec theta sin theta" "[because sintheta =(c)/(PQ)]`
`therefore h=(c)/(2) (cot theta +sqrt(3))`(1)
Now, `k=RL=RQsin(180^@-60^@-theta)`
`=PQ{sintheta xx(1)/(2)+cos theta xxsqrt3/(2)}`
`=c cosec theta{(1)/(2)+sintheta xxsqrt3/(2)costheta}`
`therefore" "k=(c)/(2)(1+sqrt3cottheta)`
Eliminating `cot theta` from (1) and (2) , we get `k=sqrt3h-c`.
Therefore the required locus is `y=sqrt3x-c`, a straight line.