$\text {The cartesian co-ordinates of the point whose polar co-ordinates are }\left(\frac{1}{2},…

$\text {The cartesian co-ordinates of the point whose polar co-ordinates are }\left(\frac{1}{2}, 120^{\circ}\right) \text { are }$
  1. $\left(\frac{1}{4}, \frac{-\sqrt{3}}{4}\right)$
  2. $\left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$
  3. $\left(\frac{-1}{4}, \frac{-\sqrt{3}}{4}\right)$
  4. $\left(\frac{-1}{4}, \frac{\sqrt{3}}{4}\right)$

Solution

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 $\mathrm{P}\left(-\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$

Asked in: MHT CET 2020 (13 Oct Shift 2)

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