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← IIT-JEE 2008 · Paper 1
MathematicsSingle correct+4 / -1Question 16 of 69

IIT-JEE 2008 · Paper 1, Mathematics question 16

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Paragraph for Question Nos. 15 to 17

A circle CC of radius 1 is inscribed in an equilateral triangle PQRPQR. The points of contact of CC with the sides PQPQ, QRQR, RPRP are DD, EE, FF, respectively. The line PQPQ is given by the equation 3x+y−6=0\sqrt3x+y-6=0 and the point DD is (332,32)\left(\frac{3\sqrt3}2,\frac32\right). Further, it is given that the origin and the centre of CC are on the same side of the line PQPQ.

Points EE and FF are given by

Choose one option
Answer and worked solution

Official answer: A

Points of contact of an equilateral triangle's incircle are 120° apart round the centre O(3,1)O(\sqrt3,1).

OD→=(32,12)\overrightarrow{OD}=\left(\frac{\sqrt3}2,\frac12\right) (direction 30°). Rotating by ±120° gives directions 150° and 270°:

  • O+(−32,12)=(32,32)O+\left(-\frac{\sqrt3}2,\frac12\right)=\left(\frac{\sqrt3}2,\frac32\right);
  • O+(0,−1)=(3,0)O+(0,-1)=(\sqrt3,0).

These are E and F.

(A)