Distance Between Two Points: From Axes to Diagonals

On previous pages you found distances between points whenever the segment between them was parallel to one of the axes. The rule was simple subtraction:

• Two points $(x_1, y)$ and $(x_2, y)$ on the same horizontal line (same y-coordinate): the distance is $|x_2 - x_1|$.
• Two points $(x, y_1)$ and $(x, y_2)$ on the same vertical line (same x-coordinate): the distance is $|y_2 - y_1|$.

For example, in Reiaan's room (Page 5), the room door $D_1(7.5, 0)$ to $R_1(11.5, 0)$ is horizontal — its width is $|11.5 - 7.5| = 4$ ft. Easy.

But the segment from $(2, 5)$ to $(7, 1)$ is not parallel to either axis. Subtracting the x-coordinates gives 5, subtracting the y-coordinates gives 4 — neither of these is the distance between the two points. They are something else: the legs of a right triangle whose hypotenuse is the distance we want.

This is the moment Baudhāyana's theorem becomes a tool of coordinate geometry.

Consider three points in Quadrant I:

• $A = (3, 4)$
• $D = (7, 1)$
• $M = (9, 6)$

Draw segments $AD$, $DM$, $MA$ on graph paper. The result is a triangle entirely inside Quadrant I. Triangle ADM is acute-angled (all three angles are less than 90°). Now we ask the obvious question: how long are its sides $AD$, $DM$, $MA$?

None of these segments is parallel to an axis. Plain subtraction will not give us any of the three side lengths directly. But we can build a right triangle for each side, where the side is the hypotenuse and the legs are parallel to the axes.

Focus on side $AD$ first. The endpoints are $A(3, 4)$ and $D(7, 1)$. To find $AD$:

1. Drop a vertical line from $A$ down to the level of $D$ (i.e. to $y = 1$). Call the foot $C$. Then $C = (3, 1)$.
2. Now we have three points: $A(3, 4)$, $C(3, 1)$, $D(7, 1)$. Notice that $\angle ACD = 90°$ — $C$ is directly below $A$ and directly left of $D$, so the segment $AC$ is vertical and the segment $CD$ is horizontal. Triangle $ACD$ is right-angled at $C$.
3. Now compute the two legs by simple subtraction:
   • $CD$ = horizontal distance = $|7 - 3| = 4$
   • $AC$ = vertical distance = $|4 - 1| = 3$
4. By Baudhāyana–Pythagoras: $AD^2 = AC^2 + CD^2 = 3^2 + 4^2 = 9 + 16 = 25$, so $AD = \sqrt{25} = 5$.

The diagonal distance from $A(3, 4)$ to $D(7, 1)$ is exactly 5 units. This is the same 3-4-5 right triangle Baudhāyana would have used to lay out a perfect right angle on the ground 2,800 years ago — and it works because of the same theorem he proved.

The pattern is so important it deserves a name. The distance between two points in a plane is the hypotenuse of the right triangle whose legs are parallel to the axes.