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IIT JEE Advanced 2008 Paper 1

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A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation 3x + 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 C are on the same side of the line PQ.

Section: Mathematics

Question : 65 of 69

Marks: +1, -0

Points E and F are given by

[JEE Adv 2008 P1]

$\left(\frac{\sqrt{3}}{2},\frac{3}{2}\right)$ , $(\sqrt{3},0)$
$\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$ , $(\sqrt{3},0)$
$\left(\frac{\sqrt{3}}{2},\frac{3}{2}\right)$ , $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$
$\left(\frac{3}{2},\frac{\sqrt{3}}{2}\right)$ , $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$

Validate

Solution:

Since slope of the line PQ =

-\sqrt{3}

, PQ makes 120° angle with x-axis.

Therefore, side PR lies along x-axis. Hence,

F =

(\sqrt{3},0)

Now, the equation of the line segment CE is

\frac{x-\sqrt{3}}{-\frac{\sqrt{3}}{2}}

\frac{y-1}{\frac{1}{2}}

= 1

⇒ E =

\left(\frac{\sqrt{3}}{2},\frac{3}{2}\right)

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