Q40 / 69 · JEE Advanced 2008 (Paper 1) · Mathematics

Paragraph: A circle $C$ of radius 1 is inscribed in an equilateral $\triangle P Q R$. The points of contact of $C$ with the sides $P Q, Q R, R P$ are $D, E, F$ respectively. The line $P Q$ is given by the equation $\sqrt{3} x+y-6=0$ and the point $D$ is $\left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right)$. Further, it is given that the origin and the centre of $C$ are on the same side of the line $P Q$.Question: Equations of the sides $Q R, R P$ are

  • A
    $y=\frac{2}{\sqrt{3}} x+1, y=-\frac{2}{\sqrt{3}} x-1$
  • B
    $y=\frac{1}{\sqrt{3}} x, y=0$
  • C
    $y=\frac{\sqrt{3}}{2} x+1, y=-\frac{\sqrt{3}}{2} x-1$
  • Correct answer
    D
    $y=\sqrt{3} x, y=0$

Solution

Clearly, point $E$ and $F$ satisfy the equations given in option (d).

Hence, the correct answer is option: 'D'.

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