Both polar and rectangular coordinate systems are a process of defining any point in a graph in a unique way. By both of these systems an ordered pair consisting of two elements, uniquely define any point on the plane.

**Rectangular coordinates** are coordinates represented in terms of x and y. The full plane is divided into 4 quadrants by two axes known as x axis or horizontal axis and y axis or vertical axis. The first quadrant consists of both x and y positive, and second quadrant x -ve, y +ve, iii quadrant, x -ve, y -ve and iv quadrant x +ve and y -ve.

We can adopt any scaling factor for x and y separately depending on the magnitutdes in the data. Thus any ordered pair (x, y) can be plotted in a grraph with appropriate scale for values of x ranging from - infinity to + infinity.

The rectangular or cartesian coordinates is very famous because of its simplicity and usage by maximum people.

The distance formula, the negative signs, location, mid point formula and all easy to locate in cartesian coordinates than any other coordinates. Also calculations are simple as in decimal system 10 partitions will be there in each grid.

Using Cartesian Coordinates you mark a point by **how far along** and **how far up** it is:

Using Polar Coordinates you mark a point by **how far away**, and **what angle ** it is:

To convert from one to the other, you need to solve the triangle:

Below are some examples based on rectangular coordinates**1. A circle has equation x**^{2}**+y**^{2}** =9 in rectangular coordinates. Find the equation of the **circle in polar coordinates**.**

**Solution: **We know that r^{2} =x^{2}+y^{2} So here r^{2} =9 or r = 3.

The equation of the circle in polar coordinates is (rcosθ,rsinθ) for 0 ≤ θ≤2π

**2. Find the mid point of (1, 2) and (3, 4) and express it in polar coordinates **

**Solution: **Mid point of (1,2) and (3,4) = (1+3/2,2+4/2) =(2,3)

x =2 y =3 Now x^{2}+y^{2} = 4+9 =13 =r^{2}

θ = tan inverse of y/x =3/2 =1.5

So mid point in polar coordinates is ( √13,tan^{-}^{1}1.5)

**3. Find the equation of this hyperbola in rectangular coordinates: (acos?,bsin?)**

**Solution: ** We know that for functrions, cos^{2}θ+sin^{2}θ=1.

Hence we can write here x^{2}/a^{2} +y^{2}/b^{2} =1 which is the equation of an ellipse.

**4. Find the circle with centre (2,4) with diameter 6 in rectangular coordinates then convert to polar coordinates.**

**Solution: ** Te equation of the circle is (x-2)^{2}+(y-4)^{2} = 9

In polar coordinates we know x =rcosθ and y = rsinθ

So we have here x =2+rcosθ and y = 4+rsinθ

For more help students can connect to an online tutor and gain free help in understanding the concept of converting polar and rectangular coordinates.

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