# How do you graph the point E(2,30^o)?

Jan 8, 2017

In cartesian form, it is $\left(2 \cos {30}^{o} , 2 \sin {30}^{o}\right) = \left(\sqrt{3} , 1\right)$
The first graph is for dotting the point. In the second, it is the intersection in ${Q}_{1}$ of $r = 2 \mathmr{and} \theta = {30}^{o}$.

#### Explanation:

graph{(x-1.732)^2+(y-1)^2-10^(-4)=0x^2 [-3, 3, -1.5, 1.5]}

For dotting the point by glow of pixels, use a circle with center at the

point and radius suitably quite small., say 0.01. This would enable a

few pixels (3 or 4 in two rows or 7 in three rows ) to glow, at the

point.

Here, the equation of the point circle is

${\left(x - \sqrt{3}\right)}^{2} + {\left(y - 1\right)}^{2} = {10}^{- 4}$.

Alternative method that suits polar frame:

I have used the equation $r = \sqrt{{x}^{2} + {y}^{2}} = 2$, for the circle, and

the equation $y = \frac{x}{\sqrt{3}}$, for $\theta = {30}^{o}$.

The intersection in ${Q}_{2}$ is $\left(2 , {30}^{o}\right)$, in polar form.

graph{(x^2+y^2-4)(1.732y-(x+|x|)/2)=0 [-6, 6, -3, 3]}

For locating $\left(a , \alpha\right)$ in polar form, you can mark the point as the

intersection of

the circle $r = a$ and the radial ( half ) line

$\theta = \alpha$.