General formula for the coordinates of points in an isosceles triangle

Question

I have the following isosceles triangle

If I know the coordinates of points A and B, and the length of the line CC', how can I determine the coordinates of points C and C'? Point A bisects CC'.

(image stolen from Finding coordinates for 3rd point in isosceles right triangle, knowing 2 other points and lengths of sides)

Answer 1

Here is a solution that does not necessitate trigonometry:

Let $r = $ distance AC.

The solution is given by

$$\tag{1}\vec{AC}=r\vec{v} \ \ \ \ \ \iff \ \ \ \ \ C=A+r\vec{v},$$

where $\vec{v}$ is a unit (length 1) vector orthogonal to $\vec{AB}$.

Let $n=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$ be the norm of $\vec{AB}$.

Expressing (1) with coordinates (using the fact that vector $\binom{-b}{a}$ is directly orthogonal to $\binom{a}{b}$):

$$\binom{x_C}{y_C}=\binom{x_A-\tfrac{r}{n}(y_B-y_A)}{y_A+\tfrac{r}{n}(x_B-x_A)},$$

and, in a symmetrical way:

$$\binom{x_{C'}}{y_{C'}}=\binom{x_A+\tfrac{r}{n}(y_B-y_A)}{y_A-\tfrac{r}{n}(x_B-x_A)}.$$

Answer 2

You can get parametric coordinates of a point $(x,y)$ at $r$ distance from a fixed point $(x_1,y_1)$ on a line with inclination from positive $x$-axis $\theta$

$$x=x_1+r \cos \theta~ ;~ y=y_1+r \cos \theta$$

You can get $\theta$ since CC' is perpendicular to $AB$ therefore $m_{AB} \cdot m_{CC'} =-1$ and you know $r = \dfrac{CC'}{2}$

Hence, coordinates of C and C' will be : $(x_1\pm r \cos \theta,y_1 \pm r \cos \theta)$

You may proceed now.

Hope it helps!