Let the coordinates of point A be (x1 y1), the coordinates of B be (x2 y2), and the coordinates for C, which is a right angle, be (x2 y1)
It is understood from the image that triangle ABC is a right angle triangle. AB is, in fact, the hyptenouse, and AC & BC are the corresponding sides.
The length of the long leg, side AC, is equal to the difference in the x-coordinates of points C and A.
Points C and B have the same x-coordinate, so the length of side AC is also equal to the difference in the x-coordinates of points B and A. So, AC = x2 − x1
The length of the short leg, side BC, is equal to the difference in the y-coordinates of points B and C.
Points A and C have the same y-coordinate, so the length of side BC is also equal to the difference in the y-coordinates of points B and A. So, BC = y2 − y1
Thus, from pythagorous theorum... a2 + b2 = c2
We substitute...
AB2 = AC2 + BC2
We substitute again... d2= (x2-x1)2 + (y2-y1)2
Now, we will take the square root of either side and arrive at the distance forumla... d = √(x2-x1)2 + (y2-y1)2
B.
Let the coordinates of point A be (x1 y1 z1), of B be (x2 y2 z1), C be (x3 y3 z1), D be (x2 y2 z2)
This derivation is two step process.
First, we will find AB and, using it, we will find AD
It is understood from the image that triangle ABC is a right angle triangle. AB is, in fact, the hyptenouse, and AC & BC are the corresponding sides
Let's suppose that AC is actually along the x-axis while BC is along the y-axis
Now triangle ABD is also right angled triangle with hypotenuse AD
From above, we know that AB = √(x2-x1)2 + (y2-y1)2
The length of segment BD is the difference in the z-coordinates between points B and D, which is also the difference in the z-coordinates of points A and D. So, BD = z2 − z1
Thus, from the Pythagorean theorem... a2 + b2 = c2
We substitute...
AD2 = AB2 + BD2
We substitute again... d2 = (√ (x2 – x1)2 + (y2 – y1)2)2 + (z2 - z1)2
Then we simplify... d2= (x2-x1)2 + (y2-y1)2 + (z2-z1)2
Now, we will take the square root of either side and arrive at the distance forumla... d = √(x2-x1)2 + (y2-y1)2 + (z2-z1)2
C.
The geometric illustration was built by first learning how each of the buttons on the toolbar to the left worked. I then began with a clean file and drew a single line segment. From there, I clicked on the line segment and went up to the "Construct" drop down menu and selected "Perpendicular Line." In this way, I continued to create more line segments, and eventually, the 3D image. The perpendicular lines allowed me to change the shape of the box without distorting it. I also hid the perpendicular lines to avoid confusion. Line segments were drawn for line AD, DB, and AB, given a color, as well as the "dotted" and "dash" effect, by right clicking on each line. I right clicked on each line segment that needed measurement, AD, DB, and AB, respectively, and selected the "Length" option in order to display the length of each line.
Finally, I took snapshots of the geometric figure, utilizing the Snipping Tool on my Windows 7 computer, in different box shapes by dragging the vertices.