# How to find the third coordinate of a right triangle given 2 coordinates and lengths of each side

 p2
|\
|b\
|  \
A|   \C
|    \
|c___a\
p1  B   p3


If given point p1 & p2, side A & B how would you find point p3? I know given this information you can find side C and all of the interior angles.

side C:
C^2 = A^2 + B^2

angle c = 90
angle a = A/SIN(a) = C/SIN(c)
angle b = 180 - (a+c)


But after this, I am trying to find point p3 and I am not sure what direction to take. Any help would be appreciated.

Edit: The triangle will not necessarily be facing upwards along an axis, it will be rotated at angles depending on exterior variables such as position of a mouse on the computer screen.

• You know the length of side $B$, and it seems one leg of your right triangle is horizontal. Thus, just add that length to the $x$-coordinate of p1... Commented Sep 15, 2011 at 16:47
• the triangle is going to be rotated at random angles that solution wont work. Commented Sep 15, 2011 at 16:49
• Then rotate the coordinates such that side $B$ is horizontal. You know the slope of side $A$, you can then derive the appropriate rotation matrix... Commented Sep 15, 2011 at 16:51
• Knowing points p1 and p2 you can find the line between them. You need the perpendicular to this line through point p1 and distance B along it. You may not know which direction to take, because given the information you have presented you can take either direction on the perpendicular. Commented Sep 15, 2011 at 16:59

Let the coordinates of $p_n$ be $(x_n,y_n)$. Then the slope of $A$ is $m_A=\frac{y_2-y_1}{x_2-x_1}$. The slope of $B$ is $m_B=\frac{-1}{m_A}=\frac{x_1-x_2}{y_2-y_1}$. Then $p_3=p_1\pm B(\frac{1}{\sqrt{1+m_B^2}},\frac{m_B}{\sqrt{1+m_B^2}})$ where the sign ambiguity corresponds to two orientations of the triangle. I have ignored issues when the sides are vertical or horizontal, which can lead to division by zero
• @RossMillikan Could you comment on how you go from the slope of $A$ and $B$ to $p3$? Commented Sep 6, 2014 at 17:20
• @nachocab: I got $m_B$ because it is perpendicular to $A$ and so the slopes are negative reciprocals. To get $p_3$ I made a unit vector with that slope and multiplied by the length, which is $B$ Commented Sep 6, 2014 at 17:23
• @RossMillikan ok, then I guess what I don't understand is where that unit vector comes from. If the vector is $(1,m_B)$ shouldn't the magnitude be $\sqrt{1+m_B^2}$? Also, does this method work for any triangle, or just right triangles? Commented Sep 6, 2014 at 18:22
• @nachocab: you are correct there should be a square root. I have fixed it. Thanks. The only place I used the right triangle was getting the slope of $B$ from the slope of $A$. If you get the slope of $B$ some other way, you can use the unit vector and length like this. Commented Sep 6, 2014 at 20:50
If your triangle is in space, the given information doesn't determine it yet, because given any such triangle, rotating it around the line joining $p1$ and $p2$ gives you another valid triangle. If your triangle is in the plane, then the information does determine p3 as long as you decide whether the order $p1$, $p2$, $p3$ is clockwise or counterclockwise. Let's say then, the triangle is in the plane and, as shown in your neat ASCII picture, $p1$, $p2$, $p3$ is clockwise.
If you know about complex numbers, then $p3-p1 = -iB/A(p2-p1)$ (because multiplication by $B/A$ changes a length $A$ vector into a length $B$ vector, and multiplication by $-i$ rotates by $90^\circ$ clockwise).
If you don't want to use complex numbers, then say $p1=(x1, y1)$, $p2=(x2, y2)$ and $p3=(x3, y3)$. Since $p1p2p3$ is a right angle, the slopes of $p1p2$ and $p1p3$ multiply to $-1$, that gives you one equation for $x3$ and $y3$. Additionally, you know that the distance from $p1$ to $p3$ is B, which gives you a second equation for $x3$ and $y3$. The system formed by those equations will have two solutions, one corresponding to $p1$, $p2$, $p3$ being clockwise and the other corresponding to counterclockwise.