Finding a coordinate over a right angle in a triangle where the other two coordinates are known

Question

      a
 B ------- C
   \      |
    \     |
     \    |
    c \   | b
       \  |
        \ |
         \|
          A

Alright, this is a triangle I have, and these are the things that are known: the coordinates of A and B, the lengths of a and c, as well as the fact that the angle under C is always a right angle. I need to find the coordinates of C.

If you choose to answer this, please be detailed in your explanation of the answer. Thank you.

Edit: I am using this in video game programming. The two objects are parts of a sentry gun. C is at the base of the barrel of the gun, and at a fixed distance (a) from B. B is the gimbal for the gun, and its position does not change. A represents an enemy, whose position does change, and the position of C needs to change on the gimbal (B) to continue pointing at the enemy.

I hope that made my intentions a bit clearer.

So far, all three answers are outside my scope of understanding. My knowledge is really lacking, perhaps too much for me to be able to tackle this problem. I would still like to try, however.

It would help me a lot to see this problem solved using real coordinates. Thank you for your answers so far.

Answer 1

1) Notation: $B = (x_b,y_b), A = (x_a,y_a),C = (x_c,y_c)$

2) The fact that the $\angle C$ is $90^\circ$ helps determine the value of

$b = +\sqrt{c^2 - a^2}$

3) Now you have equation of two circles centered around A and B respectively, with coordinates:

$$ C_1 : (x-x_a)^2 + (y-y_a)^2 = b^2 $$ $$ C_2 : (x-x_b)^2 + (y-y_b)^2 = a^2 $$

You need to find a valid intersection of these two circles satisfying the fact that the angle formed by $AC$ and $BC$ is right one.

I am not going past this since you need to show your efforts too. Best of Luck.

Answer 2

Vector BC

$( x_C -x_B) + i ( y_C-y_B)$

Vector CA

$( x_C -x_A) + i ( y_C- y_A)$

If they are perpendicular, vector dot product = 0

$(x_C -x_B)( x_C -x_A) + ( y_C- y_B)( y_C- y_A) = 0 \tag {1} $

Length of BC is $a$

$( x_C -x_B)^2 ++ ( y_C- y_B) ^2 = a^2 \tag{2} $

Solve these two equations to find coordinates of C.

Answer 3

If you know the coordinates of $A$ and $B$ you can find the length $c$, and since you know $a$, and that $a^2+b^2=c^2$, you can find $c$.

Look at your picture. Thinking of the line from $A$ to $B$ as a mirror, find the image of $C$ in that mirror, and you see that the distances $a$ and $b$ would be the same at that point as at the point $C$ that you've drawn, so there must be two solutions.

One way to proceed is brute force: Let $C=(c_1,c_2)$, $B=(b_1,b_2)$, $A=(a_1,a_2)$. Then \begin{align} b^2 & = (c_1-a_1)^2+(c_2-a_2)^2, \\[2pt] a^2 & = (c_1-b_1)^2+(c_2-a_2)^2. \end{align} Expanding the right sides, we get this system of quadratic equations: \begin{align} b^2 & = c_1^2 - 2c_1 a_1 + a_1^2 + c_2^2 - 2c_2 a_2 + a_2^2, \\[4pt] a^2 & = c_1^2 - 2c_1 b_1 + b_1^2 + c_2^2 - 2c_2 b_2 + b_2^2. \end{align} Subtracting lefts sides and right sides, we get $$ b^2-a^2 = 2c_1 (b_1 - a_1) + (a_1^2-b_1^2) + 2c_2(b_2-a_2) + (a_2^2-b_2^2). $$ This is linear in $c_1$ and $c_2$. If you solve it for $c_2$ in terms of $c_1$, then substitute that for $c_2$ in either of the quadratic equations above, you reduce it to a quadratic equation in $c_1$, which has two solutions.