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 |  \
A|   \C
 |    \
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.

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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... –  J. M. Sep 15 '11 at 16:47
the triangle is going to be rotated at random angles that solution wont work. –  Hussein Sabbagh Sep 15 '11 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... –  J. M. Sep 15 '11 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. –  Mark Bennet Sep 15 '11 at 16:59

2 Answers 2

up vote 6 down vote accepted

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

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+1 correct answer. –  Jim Thio Jan 27 '12 at 8:14
@RossMillikan Could you comment on how you go from the slope of $A$ and $B$ to $p3$? –  nachocab Sep 6 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 $B$ –  Ross Millikan Sep 6 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? –  nachocab Sep 6 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. –  Ross Millikan Sep 6 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 degrees 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.

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