# How do you convert (-4, 3) into polar coordinates?

Jan 9, 2016

If $\left(a , b\right)$ is a are the coordinates of a point in Cartesian Plane, $u$ is its magnitude and $\alpha$ is its angle then $\left(a , b\right)$ in Polar Form is written as $\left(u , \alpha\right)$.
Magnitude of a cartesian coordinates $\left(a , b\right)$ is given by$\sqrt{{a}^{2} + {b}^{2}}$ and its angle is given by ${\tan}^{-} 1 \left(\frac{b}{a}\right)$

Let $r$ be the magnitude of $\left(- 4 , 3\right)$ and $\theta$ be its angle.
Magnitude of (-4,3)=sqrt(-4)^2+3^2)=sqrt(16+9)=sqrt25=5=r
Angle of $\left(- 4 , 3\right) = T a {n}^{-} 1 \left(\frac{3}{-} 4\right) = T a {n}^{-} 1 \left(- \frac{3}{4}\right) = - 36.869$ degree

$\implies$ Angle of $\left(- 4 , 3\right) = - 36.869$ degree

But since the point is in second quadrant so we have to add $180$ degree which will give us the angle.

$\implies$ Angle of $\left(- 4 , 3\right) = - 36.869 + 180 = 143.131$

$\implies$ Angle of $\left(- 4 , 3\right) = 143.131 = \theta$

$\implies \left(- 4 , 3\right) = \left(r , \theta\right) = \left(5 , 143.131\right)$
$\implies \left(- 4 , 3\right) = \left(5 , 143.131\right)$
Note that the angle is given in degree measure.

Jan 9, 2016

Given that a point $\textcolor{b r o w n}{P \to \left(x , y\right) \to \left(- 4 , 3\right) \text{ Cartesian}}$

Then $\textcolor{b l u e}{P \to \left(5 , {143.13}^{o}\right) \text{ Polar }}$ to 2 decimal places

#### Explanation:

This is not a polar graph!!! $\textcolor{b l u e}{\text{Where it all comes from}}$

We are give the coordinates of (-4,3)
Suppose we viewed this in the context of Cartesian form and use $y = m x + c$

Then $c = 0$ and $m = \frac{y}{x} = - \frac{3}{4}$

So we would have $y = - \frac{3}{4} x$

Suppose the graph was only plotted over the range $x \to \left(0 , - 4\right)$

Then the above graph would not be continuous but be a line from
$\left(0 , 0\right) \to \left(3 , - 4\right)$

All we need now is the angle that that line makes to the x-axis and the length of that line.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Finding the Polar r value}}$

$\text{Length } = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{3}^{2} + {4}^{2}} = 5$

$\textcolor{b l u e}{\text{So the polar } r = 5}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Finding the Polar "theta" value}}$
The Polar angle $\theta$ is measured from the positive x-axis counterclockwise.

Let the angle from the line to the negative x-axis be $\phi$

Then $\phi = {\tan}^{- 1} \left(\frac{y}{x}\right) = {\tan}^{- 1} \left(\frac{3}{4}\right)$

But $\textcolor{w h i t e}{. .} \phi + \theta = 180$

so $\textcolor{w h i t e}{. .} \theta = 180 - {\tan}^{- 1} \left(\frac{3}{4}\right)$

color(blue)( theta = 143.13 to 2 decimal places
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting at all together}}$

Given that a point $P \to \left(x , y\right) \to \left(- 4 , 3\right) \text{Cartesian}$

Then $P \to \left(5 , {143.13}^{o}\right) \text{ Polar }$ to 2 decimal places