The program will then do two things:

The first is to output which quadrant the point $(x,y)$ is in$,$ when treated as a point in $2$D space. If

the point is not in any quadrant, the program should say that the point lies on the axes.

The second is to output in the point is within $3$ units of the origin.

Recall that the quadrants look like this:

Your grade will be based on good programming technique $($comments$,$ variable names, assumptions,

etc.$),$ correct outputs $($correct mathematical computations$),$ and correct program output $($good user

interface$/$explanations$.)$

As part of your upload include your java file and a picture of the programming being run on your

machine $($a screen capture $($like the example given$)$ is fine.$)$

C:$\backslash$Java$>$java CSCI$1301$ProgramHW$6$

welcome to the quadrant finder.

Enter in your $x$ value: $10$

Enter in your y value: $2$

The point $(10\mathrm{.}0,2\mathrm{.}0)$ is in the first quadrant.

The point $(10\mathrm{.}0,2\mathrm{.}0)$ is at least $3$ units away from the origin

C:$\backslash$Java$>$java CSCI$1301$ProgramHW$6$

Welcome to the quadrant finder.

Enter in your $x$ value: $0$

Enter in your y value: $-3\mathrm{.}2$

The point $(0\mathrm{.}0,-3\mathrm{.}2)$ lies on the axes.

The point $(0\mathrm{.}0,-3\mathrm{.}2)$ is at least $3$ units away from the origin

C:$\backslash$Java$>$java CSCI$1301$ProgramHW$6$

Welcome to the quadrant finder.

Enter in your $x$ value: $-1\mathrm{.}5$

Enter in your y value: $1$

The point $(-1\mathrm{.}5,1\mathrm{.}0)$ is in the second quadrant.

The point $(-1\mathrm{.}5,1\mathrm{.}0)$ is no more than $3$ units away from the origin