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 answerD$y=\sqrt{3} x, y=0$
Solution
Clearly, point $E$ and $F$ satisfy the equations given in option (d).
Hence, the correct answer is option: 'D'.