Question

How do you find the rectangular coordinates, given the polar coordinates $( - 1, - \dfrac{\pi }{6})$ ?

Answer 1

Hint : In this question, we are given the polar coordinates. A polar coordinate is of the form $(r,\theta )$ , so we get the values of r and $\theta$ , and we have to convert the given polar coordinates into rectangular coordinates, it means that we have to find the value of x and y. We know that $x = r\cos \theta$ and $y = r\sin \theta$ , using the value of r and $\theta$ we will find the value of both x and y and thus get the rectangular coordinates.

Complete step by step solution:
We are given the polar coordinates are $( - 1, - \dfrac{\pi }{6})$ so we get $r = - 1$ and $\theta = - \dfrac{\pi }{6}$
We know that
$x = r\cos \theta \\\ \Rightarrow x = - 1\cos ( - \dfrac{\pi }{6}) \;$
We know $\cos ( - x) = \cos x$
$\Rightarrow x = - \cos \dfrac{\pi }{6} \\\ \Rightarrow x = - \dfrac{{\sqrt 3 }}{2} \;$
$y = r\sin \theta \\\ \Rightarrow y = - 1\sin ( - \dfrac{\pi }{6}) \;$
We know that $\sin ( - x) = - \sin x$
$\Rightarrow y = - ( - \sin \dfrac{\pi }{6}) \\\ \Rightarrow y = \dfrac{1}{2} \;$
Hence when the polar coordinates are $( - 1, - \dfrac{\pi }{6})$ , the rectangular coordinates are $( - \dfrac{{\sqrt 3 }}{2},\dfrac{1}{2})$ .
So, the correct answer is “$( - \dfrac{{\sqrt 3 }}{2},\dfrac{1}{2})$ ”.

Note : The rectangular coordinate system is of the form $(x,y)$ and is the most commonly used coordinate system, where x is the distance of this point from the y-axis and y is the distance of the point from the x-axis. The polar coordinate system is of the form $(r,\theta )$ where r is the distance of the point from the origin and $\theta$ is the counter-clockwise angle between the line joining the point and the origin and the x-axis. Thus, we get a right-angled triangle formed by x, y and r, where r is the hypotenuse, x is the base and y is the height of the triangle, so by Pythagoras theorem, we have - ${x^2} + {y^2} = {r^2}$ and by trigonometry we have –
$\cos \theta = \dfrac{{base}}{{hypotenuse}} = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }} = \dfrac{x}{r} \\\ \Rightarrow x = r\cos \theta \;$
And similarly $y = r\sin \theta$