# How do you graph the point B(2.5, -pi/6)?

Apr 1, 2018

See below.

#### Explanation:

One way of simplifying it would be to transform the Polar coordinates into Cartesian coordinates.

In order to do that, let's take a general point $P \left(r , \theta\right)$ in the Polar system. If $O$ is the origin of the system and $P '$ is the intersection of the line going through $P$ parallel to the y-axis and the x-axis, then :

In the right angled triangle $P O P '$, we know the hypotenuse is $r$ and one of the angles of $\theta$. By this knowledge, we can deduce that, if we would transform the point P into Cartesian coordinates, the following relations take place:

${r}^{2} = {x}^{2} + {y}^{2}$
$x = r \cos \theta$
$y = r \sin \theta$,

where $x$ and $y$ are the Cartesian coordinates of $P$.

In our case:

$\theta = - \frac{\pi}{6}$
$r = \frac{5}{2}$

=> color(red)x = 5/2 * cos(-pi/6) = 5/2 * sqrt(3)/2 = color(red)((5sqrt(3))/4
=> color(red)y = 5/2 * sin(-pi/6) = - 5/2 * 1/2 = color(red)(-5/4.

Therefore, the point $B$ has Cartesian coordinates $\left(\frac{5 \sqrt{3}}{4} , - \frac{5}{4}\right)$.

It should be fairly easy to graph $B$ now.