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JEE Advance - Mathematics (2008 - Paper 1 Offline - No. 16)

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 $$\sqrt 3 x\, + \,y\, - \,6 = 0$$ and the point D is $$\left( {{{3\,\sqrt 3 } \over 2},\,{3 \over 2}} \right)$$. Further, it is given that the origin and the centre of C are on the same side of the line PQ.
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 $$\sqrt 3 x\, + \,y\, - \,6 = 0$$ and the point D is $$\left( {{{3\,\sqrt 3 } \over 2},\,{3 \over 2}} \right)$$. Further, it is given that the origin and the centre of C are on the same side of the line PQ.
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 $$\sqrt 3 x\, + \,y\, - \,6 = 0$$ and the point D is $$\left( {{{3\,\sqrt 3 } \over 2},\,{3 \over 2}} \right)$$. Further, it is given that the origin and the centre of C are on the same side of the line PQ.
Points E and F are given by
$$\left( {{{\,\sqrt 3 } \over 2},\,{3 \over 2}} \right),\,\left( {\sqrt 3 ,\,0} \right)$$
$$\left( {{{\,\sqrt 3 } \over 2},\,{1 \over 2}} \right),\,\left( {\sqrt 3 ,\,0} \right)$$
$$\left( {{{\,\sqrt 3 } \over 2},\,{3 \over 2}} \right),\,\left( {{{\,\sqrt 3 } \over 2},\,{1 \over 2}} \right)$$
$$\left( {{{\,3} \over 2},\,{{\sqrt 3 } \over 2}} \right),\,\left( {{{\,\sqrt 3 } \over 2},\,{1 \over 2}} \right)$$

Explanation

Slope of line $$PQ = - \sqrt 3 $$

Therefore, PQ make 120$$^\circ$$ angle with x-axis. So, side PR lies along X-axis.

Therefore, $$F = \left( {\sqrt 3 ,0} \right)$$

Now, equation CE is $${{x - \sqrt 3 } \over {{{ - \sqrt 3 } \over 2}}} = {{y - 1} \over {{1 \over 2}}} = 1$$

$$E = \left( {{{\sqrt 3 } \over 2},{3 \over 2}} \right)$$

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