What is the locus of the third vertex of a triangle when its first two vertices are moved along the x and y axes respectively?

So, we start with triangles ABC and DEF. Vertices A and D will move vertically along the y-axis and vertices B and E will move horizontally along the x-axis. We are interested in the path or the locus of vertices C and F as the other two pairs of vertices move along the axis.

To reference for the remainder of this write-up,

hereis my GSP file with my investigations.My first obstacle for exploring this problem was the mere construction of the diagram. I found it difficult to construct the figure so that it would animate and perform like I wanted it to. I wanted to be able to control the length of AB with a control to the side of the diagram and to ensure that the triangle remained rigid when the vertices were animated along the axes. This required a series of trials. In the end, I found the secret to be using circles to mark the distances. The steps for constructing this figure are below.

1. Construct a segment to be used as a measuring tool (a radius). This will be the length of AB.

2. Plot a point A on the y-axis. Mark the length of our measuring segment and then point A. Make a circle with center A and radius of the marked length.

3. Label the intersection of the circle with the x-axis as B.

4. Construct a line through A and B and rotate that by a chosen angle.

5. Make a circle with center A and a chosen radius. Mark the intersection of the circle and the rotated line through B as C.

6. Complete the triangle, ABC.

7. Make the triangle in the opposite quadrant. (can be done in various methods, i.e. rotations, translations, intersections of the circles and axes, etc)

8. Animate point A along the axis. (Everything should move based on that one point. The triangle side lengths should remain constant.)

9. Trace the two points not on the axes, C and F on my diagram.

10. Construct the locus of this path. This requires two steps. First, select A then C and construct locus. This will give you one half of the path, the path taken by C. Second, select A then F and construct locus. This will complete the path.

This process should yield the following.

From this diagram, my speculation was that the locus of these vertices was an ellipse. But how can I be sure? I thought that if I could find the equation of the ellipse then that would help to confirm my visual observations.

First, it is obvious that this is not an ellipse as it is normally presented. It is tilted so that it is not in standard form. So, we must use alternate methods of writing the formula. I will use parametric equations to define this ellipse. How do I do this? First, I will need to add a few more elements to my diagram, as seen below.

First, I will need to construct the altitudes of the triangle from the vertices that create our locus to the opposite side of the triangle The altitude will be labeled h. Now, where the altitude hits the opposite side (at the foot of the altitude), the segments AB and ED are divided. The length from A to the foot of the altitude (U) is a. The length from the foot of the altitude (U) to B is b. Both a and b will be critical in our formula. One interesting discovery at this point is to trace the foot of the altitude in both triangles. This gives us another ellipse! This time it is standard form. We can use the length of the longer and shorter axes to find the formula for this inner ellipse. Even more interesting, the longer and shorter axes are the same as the length of a and b, on the side of our triangle So the equation of the inner ellipse is: where t is from 0 to 2π.

Now, there is a connection between this ellipse and the ellipse of interest (the locus of the triangle vertices). We want to take that ellipse in the middle and tilt it by some amount and expand it by some amount so that we end up with the outer ellipse. The specific vector that we are translating or expanding this vector by is h, the altitude of the triangle Now, the equation of the ellipse that is the locus of the triangle vertices is:

where t ranges from 0 to 2π.Note here that t is a variable of angle measure. It is a variable, just like x and y in other equations. Therefore, we will not input any specific number for that. Also, note that I am not using any actual numbers. This is because the other variables, a, b, and h, depend on my length that I chose for AB on the slider tool. However for the sake of clarity, I will write the explicit equation for the particular conditions in the diagram above.

Now, what would happen if the triangles we are rotating or moving along the axes was a right triangle?

First of all, the construction for this situation is the same as for the previous one. However, the only difference is that your "chosen angle" must be 45 degrees so that the resulting triangle is indeed a right triangle. Again, if we animate the vertices, we can look at the path of the two vertices not on the axes. Will anything change with a right triangle?? In the following image, you can see that, indeed, it does change. Instead of the path being a curve, it is just one straight line. What is the equation of this line?

Just by looking at the graph, it appears that it might be y = x. But how can we show that?

One way to do this is by just moving around Point A by hand to get a better understanding of the situation. I tried to drag A up the y-axis in attempt to make a "square" with the right trianlge being the upper, right half of the square and the x- and y-axes completing the square. Without finding the exact place for A, I can still make a conjecture that the line is actually y = x. Why?

When I make my square, I know that the lengths of the two legs of the trianlge are equal and are the same as the distances from the origin to A and the origin to B since we created a "square". From that, I know that my slope of the diagonal of the square (the locus of the third vertices) will be 1. That is becasue the change in y from the origin to the third vertex will be the same as the change in x from the origin to the third vertex. Since these changes will be the same, the slope will be 1. And it is also obvious that the intercept of our line is 0.

This leads me to another approach to verifying that the equation is y = x. Now, let me drag A to the origin.When this is the case, the hypotenuse is along the x-axis.

Since these triangle are right isosceles triangles by construction, the base angles are 45 degrees. Thus, the side of the trianlge AC makes a 45 degree angle with the axes. This information, again, leads us to the idea that the slope is 1. And our y-intercept is still 0 so again we have y = x.