3
$\begingroup$

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

enter image description here

$\endgroup$
4
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Thanks, but it doesn't work. If I give a simple example A [1,1], B [11,11], C [5,1]. Here it is easy to deduce that it is D [5,5], but using your formula, I figured that y4=6.3294. What am I doing wrong, please? $\endgroup$
    – yoda666
    yesterday
2
$\begingroup$

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:

  1. $(x_1-x_3)(x_4-x_3)+(y_1-y_3)(y_4-y_3)=0$. (Because $AC$ is perpendicular to $CD$)
  2. $\frac{y_4-y_2}{y_2-y_1}=\frac{x_4-x_2}{x_2-x_1}$ (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 $x_4,y_4$

$\endgroup$
3
  • $\begingroup$ Thanks, but there's a mistake somewhere. If I give a simple example A [1,1], B [11,11], C [5,1]. Here it is easy to deduce that it is D [5,5], but using your formula, I figured that x4=5 but y4=1 and I didn't figure out where the problem is? Am I doing something wrong or is there a mistake in the formula, please? $\endgroup$
    – yoda666
    yesterday
  • $\begingroup$ @yoda666 Thanks for pointing out. Apologies for my negligence. There were some typographical errors in my answer in both equations. I have edited it. Should be fine now. Cheers! $\endgroup$
    – Shagchi
    yesterday
  • $\begingroup$ Yes, now the calculation is correct y4 = 5. Thanks again for the simple calculation. $\endgroup$
    – yoda666
    yesterday
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

You can find angle A using cosine theorem for ABC. Then use it to find length of AD from definition of cos for triangles

$\endgroup$
2
  • $\begingroup$ I don't understand how we calculate this using a cosine theorem? This is good if I know all sides, but I don't know the angles. Or when I know only one angle and two adjacent sides. How will this help me here? Plus, in a right triangle, where can I do with the Pythagorean and Sin theorems? $\endgroup$
    – yoda666
    12 hours ago
  • $\begingroup$ You know coordinates of all points so you know all the sides of ABC, I have a typo in the answer $\endgroup$
    – RiaD
    6 hours ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.