# Drawing a triangle with 2 known corners and all side lengths

Assume that there are three points $A$, $B$ and $C$.

All the pairwise distances are known $(|AB|, |AC|, |BC|)$. But none of the coordinates are known. I want to draw a triangle using those points.

Since no coordinates are known, I pick $(0,0)$ for $A$ and $(0, |AB|)$ for $B$.

Now, how can I calculate the coordinates of $C$ using these corners?

I tried:
$C_x = \dfrac{{-|AB|^{2} + {|BC|}^{2} - {|AC|}^{2}}}{2|AB|}$
$C_y = \sqrt{|AC|^2 - {C_x}^2}$

But I think that does not work.

I am using trilateration algortithm to find the realtive positions of the points. When I pick 3 points with their actual coordinates, the algorithm works perfect.

However, when I pick three points and estimate their coordinates using the formula above, what I get is something like this:

I think there is something wrong with the initial coordinates I'm picking.

• You should get 2 solutions for $C_x$. You may post your calculus so we can review it May 21, 2014 at 10:08
• @Bilou06 is the formula correct? May 21, 2014 at 10:10
• if $|BC|=0$ and $|AB|=|AC|$, you should get $B=C$, and you don't May 21, 2014 at 10:15
• First, if triangle inequality holds for your ABC (and it should!) then you get that $C_x$ is always negative. Ouch. Also, if you are not sure which side is the largest of three, then $C_x$ may lie inside AB or outside it (and obtuse-angled triangle). May 21, 2014 at 10:42
• Well, not always negative, just for most triangles where BC is not very large. For example, for BC = 1 and AC=AB=10 you get an isosceles triangle with... an obtuse angle at the base, if your formula is to be believed :) May 21, 2014 at 10:49

Pick the longest side, $c$. Draw AB of length $c$ along $x$-axis, starting from zero. Let $C_x = x$. From the right-agled triangles featuring the height of the triangle we have $$b^2 - x^2 = a^2 - (c-x)^2= a^2 - c^2 +2cx - x^2$$ $$2cx = b^2 + c^2 - a^2$$
Then the coordinates of point $C$ are $$C_x = \frac {b^2 + c^2 - a^2}{2c}$$ $$c_y=\sqrt {b^2 - C_x^2}$$
I haven't checked it thoroughly, but seems that I got it right. Oh, and of course, if it is done with given arbitrary numbers $a,b,c$ then you should first check if the triangle inequality holds or it will just going to be embarassing for you when the formulas don't work.
The point $C$ is the point of intersection of two circles $$x^2+y^2=|AC|^2$$ and $$x^2+(y-|AB|)^2=|BC|^2$$ which obviously are two $symmetric$ points.