Learn the concept of polar equation in this page with solved examples and diagrams provided for a better understanding. First let's understand what is polar coordinate system.

Two dimensional coordinate system for determining the distance from the fixed point and determining an angle from a fixed direction on a plane. This is called polar coordinate system which can be found on each point. The **pole** is the fixed point and the ray from that point with the fixed direction is the **polar axis**. The distance from the point is known as the ** radius **, and the angle is the **polar angle. **A point in the plane is addressed by the coordinates (r, theta).

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Below is provided the method to solve polar equation of line:

**To find straight line in polar coordinates**

The straight line is y = mx + c can be expressed in polar coordinates form as

y = mx + c

r sin θ = m × r cos θ + c

r(sin θ - mcos θ) = c

`r = c/(sin theta - m cos theta)`

So the straight line equation y = mx + c is represented in polar coordinates by

`r = c/ (sin theta - m cos theta)`

**Solve the polar into Cartesian form: **

** **To convert the polar straight line coordinates (r, θ) into the straight line Cartesian coordinates (x, y) with the help of formula, given below.

Here, by definition,

sin (θ) = `y/r`

cos (θ) = `x/r`

Therefore,

x= r cos (θ)

y=r sin (θ)

Here are few example problems on polar equation of line for a better understanding:

**Problem 1: **Solve the polar coordinates (5, 90°) into Cartesian coordinates

**Solution: **Change the polar into Cartesian form

Here, r = 5 and θ = 90°

x = r cos (θ)

y = r sin (θ)

x = 5 cos 90°

= 5 × 0

x = 0

y = 5 sin 90°

= 5 x 1

y = 5

Cartesian form is (0, 5)

**Problem 2:** What are each of the fourth-roots of expressed in polar form?

**Solution: ** r = ` |sqrt(3)+i|` ** **

r = `"sqrt(sqrt(3^2+1^2))" `

r = `"sqrt(3+1)"`

r = 2

Since cos α = `sqrt(3/2)` and sin α = `1/2` , α is in the first quadrant and α = 30°. Therefore, since the sine and cosine are periodic,

z = r(cos α + i sin α )

z = 2[cos ( + k. ) + i sin ( + k. )]

and applying the theorem of nth root, the first four-roots of *z* are

`2^(1/4)`

where *k* = 0, 1, 2, and 3

Thus the fourth-roots are

z_{1} = `2^(1/4)(cos 6^o + i sin 6^0)`

z_{2} = `2^(1/4)(cos 78^o + i sin 78^o)`

z_{3} = `2^(1/4)(cos150^o + isin150^o)`

z_{4} =`2^(1/4)(cos 222^o + i sin 222^o)`

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