How to find the coordinates of a point in a triangle if the other 2 points coordinates and length between them are known?

Question

I have a simple right-angle triangle $\triangle ABC$. I know the coordinates of $B$ and $C$ and the distances $AB = c$, $AC = b$ and $BC = a$. How can I find the coordinates of point $A$. For example, if $B = (4,3)$ and $C = (4,0)$ and the distance from A to B is $5$, how can I get the coordinates of $A$, which in this case $A = (0,0)$.

Edit:

The $AC$ and $BC$ Wont always be parallel to the $x$ and $y$ axis respectively

Answer 1

There is a way to calculate the coordinates of $A$ directly. But note, that there are always $2$ possible points for $A$ satisfying the given conditions (just reflect the triangle about $BC$).

• Caclulate from the coordinates of $B$ and $C$ the vector $\vec a = \vec{CB} = \begin{pmatrix}a_x \\ a_y \end{pmatrix}$.
• Now, $\begin{pmatrix}-a_y \\ a_x \end{pmatrix}$ is orthogonal to $\vec a = \vec{CB}$. If $\vec{OC}$ denotes the positional vector of $C$, the two possible positional vectors for point $A$ are

$$\vec{OA} = \vec{OC} \pm \frac ba \begin{pmatrix}-a_y \\ a_x \end{pmatrix}$$

Additional example referring to a comment:

• If $P(x,y)$ is any point in the coordinate system, then the positional vector $\vec{OP}$ is given by $\vec{OP} = \begin{pmatrix}x \\ y \end{pmatrix}$. $O$ corresponds to the point $O(0,0)$.
• $\vec a = \vec{CB}=\vec{OB}-\vec{OC} = \begin{pmatrix}4 \\ 3 \end{pmatrix}- \begin{pmatrix}4 \\ 0 \end{pmatrix} = \begin{pmatrix} 0\\ 3 \end{pmatrix}$
• An orthogonal vector to $\vec a$ with length $a$ is (using above formula) $\begin{pmatrix}-3 \\ 0 \end{pmatrix}$.
• Now, normalize this vector (scale it to a vector of length $1$) and multiply it by the length $b$: $$\frac ba\begin{pmatrix}-3 \\ 0 \end{pmatrix}= \frac 43 \begin{pmatrix}-3 \\ 0 \end{pmatrix}= \begin{pmatrix}-4 \\ 0 \end{pmatrix}$$
• Now, add or subtract this vector to the positional vector of $C$ to get the possible positional vectors of point $A$:

$$\vec{OA} = \begin{pmatrix}4 \\ 0 \end{pmatrix} \pm \begin{pmatrix}-4 \\ 0\end{pmatrix}\Rightarrow \vec{OA}=\begin{pmatrix}0 \\ 0 \end{pmatrix} \mbox{ or } \vec{OA}=\begin{pmatrix}8 \\ 0 \end{pmatrix}$$

Answer 2

Hint:

You can compute the distance $BC = \sqrt{\left(x_c - x_b\right)^2 + \left(y_c - y_b\right)^2}$, where $B = \left(x_b, y_b\right)$, and $C = \left(x_c, y_c\right)$. Because you now know $AB$ and $BC$, you can find the third length $AC$ with the help of the Pythagora's Theorem, i.e., $AC = \sqrt{AB^2 - BC^2}$. Finally, let $A = (x_a, y_a)$. Then, $$\begin{align}AC &= \sqrt{\left(x_c - x_a\right)^2 + \left(y_c - y_a\right)^2}\tag{1}\\AB &= \sqrt{\left(x_b - x_a\right)^2 + \left(y_b - y_a\right)^2}\tag{2}\end{align}$$ Solve the equations simultaneously to get the values for variables $x_a$ and $y_a$.

Answer 3

No. Since the coordinates of A,B are same, side a is parallel to y~axis.You have a right angle at C so b is likewise parallel to x-axis.

Side C is hypotenuse of length 5 and by Pythagoras thm it gives $ b=\sqrt{5^2-3^2} = 4. $

The x-coordinate of c has to be moved left by 4 units to reach A.

So coordinates of A are $$(4-4, y_C) = (0,0)$$