MHT CET · Maths · Straight Lines
Let \(P \equiv(-5,0), Q \equiv(0,0)\) and \(R \equiv(2,2 \sqrt{3})\) be three points. Then the equation of the bisector of the angle \(P Q R\) is
- A \(x-\frac{\sqrt{3}}{2} y=0\)
- B \(\frac{\sqrt{3}}{2} x-y=0\)
- C \(x+\sqrt{3} y=0\)
- D \(\sqrt{3} x+y=0\)
Answer & Solution
Correct Answer
(D) \(\sqrt{3} x+y=0\)
Step-by-step Solution
Detailed explanation
Slope of \(\mathrm{QR}=\frac{3 \sqrt{3}-0}{3-0}=\sqrt{3}\) i.e., \(\theta=60^{\circ}\)
Clearly, \(\angle \mathrm{PQR}=120^{\circ}\)
OQ is the angle bisector of the angle PQR , so line OQ makes \(120^{\circ}\) with the positive direction of X-axis.
Therefore, equation of the bisector of \(\angle \mathrm{PQR}\) is \(y=\tan 120^{\circ} x \Rightarrow y=-\sqrt{3} x \Rightarrow \sqrt{3} x+y=0\)
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