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# Transformation of Coordinates

Top
 Sub Topics The points are defined in space in the form of coordinates. The format of representing coordinates in space is known as coordinate system. There are different types of coordinate systems defined in coordinate geometry such as cartesian coordinate system, polar coordinate system, cylindrical coordinate system and spherical coordinate system. In each coordinate system, the way of representing coordinates of a point is different. The coordinates of a point can be transformed from one system into another system. The most common transformations are polar to Cartesian and Cartesian to polar which are described below in this page. Two-dimensional Cartesian representation of a point is illustrated below:Two-dimensional polar representation of a point is shown below:

## Polar Coordinates to Cartesian Coordinates

Let us consider the polar coordinates of a point are $(r,\ \theta )$. The formulas for converting polar coordinates to cartesian coordinates are:
Where, (x, y) are the coordinates of the point in cartesian coordinate system.
Let us consider one example.

### Solved Example

Question: Convert polar coordinates $(3,60^{\circ})$ of a point into Cartesian coordinates.
Solution:
Here, r = 3 and $\theta =60^{\circ}$

$x=r\cos \theta$

$x=3\cos 60^{\circ}$

$x$ = $\frac{3}{2}$

$y=r\sin \theta$

$y=3\sin 60^{\circ}$

$y$ = $\frac{3\sqrt{3}}{2}$

So, the required Cartesian coordinates are ($\frac{3}{2}$$\frac{3\sqrt{3}}{2}). ## Cartesian Coordinates to Polar Coordinates Back to Top Let us consider the Cartesian coordinates of a point are (x, y). The formulas for converting Cartesian coordinates to polar coordinates are: Where, (r,\ \theta ) are the coordinates of the point in polar coordinate system. Let us consider the same previous example. ### Solved Example Question: Convert Cartesian coordinates (\frac{3}{2}, \frac{3\sqrt{3}}{2}) of a point into polar coordinates. Solution: Here, x = \frac{3}{2} and y = \frac{3\sqrt{3}}{2} r=\sqrt{x^{2}+y^{2}} r = \sqrt{(\frac{3}{2})^{2}+(\frac{3\sqrt{3}}{2})^{2}} r = \sqrt{\frac{9}{4}+\frac{27}{4}} r = 3 \theta =\tan^{-1}$$(\frac{y}{x})$

$\theta =\tan^{-1}$$(\frac{\frac{3\sqrt{3}}{2}}{\frac{3}{2}})$

$\theta =\tan^{-1}(\sqrt{3})$

$\theta =\tan^{-1}(\tan 60^{\circ} )$

$\theta =60^{\circ}$

So, the required Cartesian coordinates are $(3,60^{\circ})$.