JEE Advanced · Mathematics · 9. Straight Lines
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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
\(y=\frac{\sqrt{3}}{2} x+1, y=-\frac{\sqrt{3}}{2} x-1\)
- D
\(y=\sqrt{3} x, y=0\)
Answer & Solution
Correct Answer
(D)
\(y=\sqrt{3} x, y=0\)
Step-by-step Solution
Detailed explanation
Clearly, point \(E\) and \(F\) satisfy the equations given in option (d).
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