Question

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

Answer 1

In cartesian form, it is `(2cos 30^o, 2 sin 30^o)=(sqrt3, 1)`
The first graph is for dotting the point. In the second, it is the intersection in `Q_1` of `r =2 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

`(x-sqrt3)^2+(y-1)^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 = x/sqrt3`, for `theta = 30^o`.

The intersection in `Q_2` is `(2, 30^o)`, in polar form.

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

For locating `(a, alpha)` in polar form, you can mark the point as the

intersection of

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

`theta = alpha`.