If we want to show a computer how to draw lines in a virtual world (using a 2-D cartesian grid), then the best tools are things called parametric equations.  When drawing a line, the 3 parameters we need are:
(1) the starting point: (x₀, y₀), (2) the direction of the line: D_x, D_y, and (3) the length of the line: S.
The “parametric equations” that draw a line are written like this:

    x = x₀ + (D_x * S)
    y = y₀ + (D_y * S)

where x and y mark the point where the “pencil” is placed on the grid when drawing the line. Notice how “symmetrical” the equations are. Can you guess what the “z” coordinate would look like for drawing lines in 3-Dimensions?

We will learn how to find the D_x, D_y direction parameters later. Notice that plugging S = 0 in the equations above sets the “pencil” at the starting point (x = x₀, y = y₀).  When S = 1, the pencil has moved a distance of 1 unit away from the starting point, a distance D_x in the x-direction and a distance D_y in the y-direction. The pencil is now at the point (x = x₀ + D_x, y = y₀ + D_y), as seen when plugging S = 1 into equations. See the diagram above. Since the “pencil” has moved a total distance of 1 unit in total, we know from the Pythagorean theorem that:

   D_x² + D_y² = 1².

To draw our first line, let us first look at the simplest example: drawing the side S₃ of the triangle in the diagram.  Note that we have changed the diagram a bit from previous posts, here we use the symbols x₃ and y₃ to specify the point P₃:(x₃, y₃) and we now name the symbols x and y to mark the point where the pencil is drawing the line on the grid.

We picked the grid so that the line segment S₃ lies along the x-axis. If we start at the point P₁:(0, 0), then the starting point parameters are: x₀ = 0 and y₀ = 0.  This line segment ends at the point P₂:(S₃, 0), where x = S₃, y = 0, and S = S₃.  Thus, plugging these values into the parametric equations gives us:

     x = S₃ = 0 + (D_x * S₃)
     y = 0 = 0 + (D_y * S₃)

It is not too hard to see that D_x = 1 and D_y = 0. And so, the complete parametric equations, with these parameters are:

     x = 0 + (1 * S) = S
     y = 0 + (0 * S) = 0

Let’s try this out. Remember, the point (x, y) determines the point where the “pencil” is drawing the pixel on the computer screen. To draw the S₃ side you start with S = 0, where the “pencil” is at point P₁:(0, 0) and then increase the length of the line, S, by one pixel length, compute x and y and draw in the next pixel at the point (x, y).  You keep on increasing S one pixel length at a time and drawing in pixels until the total length of the complete line segment is S₃, when S = S₃, you have reached the point P₂:(S₃, 0) and the S₃ line segment is complete.

Note that you could also draw the line by putting your pencil on starting point P₂ and drawing in the negative direction until you reach P₁.  Since you are drawing in the negative direction, this time the parameter x₀ = S₃ and D_x = -1. Plugging these into the parametric equation for this line segment:

   x = S₃ + (-1 * S) = S₃ – S
   y = 0

Try it out.  When S = 0, your pencil is at P₂:(S₃, 0).  As S increases, the line is growing in the negative direction until S = S₃ and then x = S₃ – S₃ = 0 at P1:(0, 0).  The same line segment has been drawn backwards.

Things get a little more interesting when finding the parameters to draw the side S₂.  Let’s start at P₁:(0, 0) and draw a line segment S₂ to P₃:(x₃, y₃).  As before, at the starting point P₁, x₀ = 0 and y₀ = 0.  Things get just a bit more interesting when trying to find the direction parameters D_x and D_y.  The parametric equations are:

     x = 0 + (D_x * S) = D_x * S
     y = 0 + (D_y * S) = D_y * S

and at the endpoint P₃, the parameters are S = S₂, x = x₃ and y = y₃. Plugging them in:

    x = x₃ = D_x * S₂  and so D_x = x₃/S₂
    y = y₃ = D_y * S₂ and…     D_y = y₃/S₂

Note that: D_x² + D_y² = x₃²/S₂² + y₃²/S₂² = (x₃² + y₃²)/ S₂² = 1, recall that x₃² + y₃² = S₂² from the Pythagorean Theorem and everything checks out.

Finally, the parametric equations can be written:

     x = (x₃/S₂) * S
     y = (y₃/S₂) * S

Now, in the last blogpost, we saw that the definition the Cos(A₁) and Sin(A₁) were:

Cos(A₁) = x₃/S₂
Sin(A₁) = y₃/S₂

These are exactly what you see in the equations above, Thus, the parametric equations for the line segment S₂ are:

       x = Cos(A₁) * S
y = Sin(A₁) * S

Where S goes from 0 to S₂. Quickly, if we want to draw this line in reverse, from P₃ to P₁, then our starting point is (x₃, y₃), moving in the negative direction, thus:

      x = x₃ – Cos(A₁) * S
y = y₃ – Sin(A₁) * S

Again S goes from 0 to S₂.

The equations above will also draw circles.  If you keep the distance (S) constant and vary the angle (A₁) from 0 to 360 degrees, the (x, y) “pencil” will sweep out all the points on a circle that are a radius S away from the center (x₀, y₀).  This is an extra bonus.

For reference, below are the general parametric equations for drawing both lines and circles:

       x = x₀ + Cos(A₁) * S
       y = y₀ + Sin(A₁) * S

If you want to draw lines, vary S.  If you want to draw circles, vary the angle A₁.

Our last task is finding the parametric equations to draw the last side, S₁.  Let’s start at the point P₂:(S₃, 0) and go to P₃:(x₃,  y₃).  And so, our starting point is x₀ = S₃ and y₀ = 0.  We just need to find the D_x and D_y parameters.  We can find these by looking at the parameters at the ending point, x = x₃ and y = y₃, and S = S₁, thus:

     x₃ = S₃ + (D_x * S₁)   thus   D_x = (x₃ – S₃)/S₁
     y₃ = 0 + (D_y * S₁)    thus   D_y = y₃/S₁

If we want to write D_x and D_y in terms of the angle A₂, then we need to find expressions for Cos(A₂) and Sin(A₂). Looking at the diagram, the triangle in the left dashed box has a base of length x₃ and the side S₂, forming the angle A₁.  In the right dashed box, the triangle has the base length S₃-x₃, with the side S₁ that form the angle A₂.  The left and right triangles face each other and are joined by the common height length y₃.  If we map out the corresponding parts of the left and right triangles:

Left –> Right
    x₃ –> (S₃-x₃)
    y₃ –> y₃
    A₁ –> A₂
    S₂ –> S₁

                                    Thus

Cos(A₁) = x₃/S₂  –> Cos(A₂) = (S₃ – x₃)/S₁
Sin(A₁) = y₃/S₂   –>  Sin(A₂) = y₃/S₁

This idea of mapping is very powerful, especially when it is used to change perspectives from one frame of reference to another.  In this case, it gives us the desired expressions that we need to draw the line segment S₁ in the perspective of the angle A₂. Below are the resulting parametric equations, with D_x and D_y as found a few paragraphs ago:

   D_x = (x₃ – S₃)/S₁ = -(S₃ – x₃)/S₁ = -Cos(A₂)
   D_y = y₃/S₁ = Sin(A₂)

Thus

     x = S₃ + (-Cos(A₂) * S) = S₃ – (Cos(A₂) * S)
     y = Sin(A₂) * S

Check it out!  After a little practice, finding the parametric equations for lines and circles becomes very easy to do, much easier than other methods.

 One huge advantage of using these parametric equations is that the parameter S in these equations is the actual distance of the point (x, y) from the starting point (x₀, y₀).  At times, this is convenient when measuring the distance between points along a given angle.  Some examples of this will be given in later posts.  It makes geometry problems so much easier.