MHT CET · Maths · Straight Lines
The cartesian co-ordinates of the point whose polar co-ordinates are \(\left(\frac{1}{2}, 120^{\circ}\right) \text { are }
\)
- A \(\left(\frac{1}{4}, \frac{-\sqrt{3}}{4}\right)\)
- B \(\left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)
- C \(\left(\frac{-1}{4}, \frac{-\sqrt{3}}{4}\right)\)
- D \(\left(\frac{-1}{4}, \frac{\sqrt{3}}{4}\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\frac{-1}{4}, \frac{\sqrt{3}}{4}\right)\)
Step-by-step Solution
Detailed explanation
Given \(\mathrm{P}(\mathrm{r}, \theta)=\left(\frac{1}{2}, 120^{\circ}\right) \Rightarrow \mathrm{r}=\frac{1}{2}, \theta=120^{\circ}\)
We have \(\mathrm{x}=\mathrm{r} \cos \theta=\frac{1}{2} \cos 120^{\circ} \quad=\frac{1}{2}\left(-\frac{1}{2}\right) \Rightarrow \mathrm{x}=-\frac{1}{4}\)
and \(y=r \sin \theta=\frac{1}{2} \sin 120^{\circ}=\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}\)
\(\therefore\) Required point is \(\mathrm{P}\left(-\frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)
We have \(\mathrm{x}=\mathrm{r} \cos \theta=\frac{1}{2} \cos 120^{\circ} \quad=\frac{1}{2}\left(-\frac{1}{2}\right) \Rightarrow \mathrm{x}=-\frac{1}{4}\)
and \(y=r \sin \theta=\frac{1}{2} \sin 120^{\circ}=\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}\)
\(\therefore\) Required point is \(\mathrm{P}\left(-\frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)
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