How to calculate the side of a right triangle from the coordinates of points and the length of one side? I have the line AB. And I need to calculate the coordinates of point D.
I know the coordinates of points A, B and C.
If I make this an imaginary right triangle, I just need to know the length of the CD line (a in the picture)
Since I can easily calculate the line length AC (d on the picture) from the coordinates, I only need the line AD to calculate the CD using the Pythagorean theorem.
I know the coordinates of points A and B, so I can easily calculate the length of line AB from this.
But how do I calculate the length of the AD line so that I can then calculate the length of the CD? Or is it possible in another way? Unfortunately, I don't know the angles either.
Please help
Thank you

 A: Let $t$ the $y$ coordinate of poinr $D$. We have by simili relations that:
$$\frac{\sqrt{(x_4-x_2)^2+(t-y_2)^2}}{\sqrt{(x_1-x_4)^2+(y_1-t)^2}}=\frac{|y_3-t|}{|t-y_2|}$$
Now, you can notice that putting $O(0,0)$ on the point $A$, the points $D$ and $C$ have the same $x$ coordinate. And the equation becomes:
$$\frac{\sqrt{(x_3-x_2)^2+(t-y_2)^2}}{\sqrt{(x_1-x_3)^2+(y_1-t)^2}}=\frac{|y_3-t|}{|t-y_2|}$$
Can you finish it from here?
A: Hint: Obtain the equation of the line (say $L$) passing through $C$ and perpendicular to the line $AC$. Calculate the point of intersection of $L$ and the line $AB$. This should give you the coordinates of $D$. In other words, if $D$ has the coordinates $(x_4,y_4)$, then you have the following constraints:

*

*$(x_1-x_3)(x_4-x_3)+(y_1-y_3)(y_4-y_3)=0$. (Because $AC$ is perpendicular to $CD$)

*$\frac{y_4-y_2}{y_2-y_1}=\frac{x_4-x_2}{x_2-x_1}$ (Because $D$ lies on $AB$. The cases where $x_1=x_2$ and/or $y_1=y_2$ can be dealt with similarly)

These can be solved to obtain $x_4,y_4$
A: This works only for Pythagorean triples or for triangles "similar" to Pythagorean triples. If $\space d \space$ is not an odd integer, round to an odd integer  greater than $\space 1 \space$and then multiply or divide all sides  (as needed) by
$\space \dfrac{d}{\text{rounded-number}}$ after the following calculations are complete.
Assuming  Pythagorean  triples where all sides are integers, we begin with the Pythagorean theorem where
$\space A^2+B^2=C^2\space $ and Euclid's formula where
$\space A=m^2-k^2 \quad B=2mk\quad C=m^2+k^2.\quad$
From your diagram we let the $x$-coordinate be $\space d=A.\quad$ For any primitiive Pythagorean triple, $\space A=2x+1, x\in\mathbb{N}\space$ and, for each $A$-value, there are
$2^{n-1}$ primitive triples where $\space n\space$ is the number of distinct prime factors of $\space A.\quad$
We can find all of these triples by solving the
$A$-function for $\space k\space$ and testing a defined range of $m$-values to see which yield integers. Note that there may be additional triples found if they are odd square multiples of primitives.
Let us begin
$$A=m^2-k^2\implies k=\sqrt{m^2-A}\\
\text{for}\qquad  \sqrt{A+1} \le m \le \frac{A+1}{2}$$
The lower limit ensures $k\in\mathbb{N}$ and the upper limit ensures $m> k$.
$$A=3\cdot 5=15\implies \sqrt{15+1}=4\le m \le \frac{15+1}{2} =8\\
\text{and we find}\quad
\sqrt{4^2-15}=1,\space \sqrt{8^2-15}=7\\
i.e. \space  m\in\{4,8\}\implies k \in\{1,7\} $$
$$F(4,1)=(15,8,17)\qquad \qquad F(8,7)=(15,112,113) $$
Here, there are $2^{2-1}=2$ triples where $a=8$ or $a=112$ and $AD$ is either $17$ or $113$.
In this case,
$\space \theta\approx 28.1^\circ\space$
0r
$\space \theta\approx 82.4^\circ\space$
If you happen to know an angle, you can find the closest matching Pythagorean triple using techiques described
here.
A: You can find angle A using cosine theorem for ABC. Then use it to find length of AD from definition of cos for triangles
