0
$\begingroup$

Suppose I have 3 points. A and B and C. I want the line segment through C, parallel to the line AB. point with given distance

I want the D with the given distance of d on the line which connects C to the AB line with the slope of 90 degrees (vertical). Basically I want to move the point C, closer to the AB line (half the distance in my case).

I know how to get a point between 2 points but I don't know how to get the point on the AB line.

Edit:

I have A, B and C points. I want the bold line below: what I need

It's at half the distance between the AB and C and has the same slope as AB.

How I solved it

I didn't quite use the distance. I got the parallel line through the C point and got the line between the 2 lines by calculating the mid point between A and A' and B and B'. Then connected the 2 new points.

$\endgroup$
  • $\begingroup$ Sorry , but your questions seems unclear for me! what do you need exactly ? equation of st line ? or just the point of the orthogonal projection of C on AB ? $\endgroup$ – Nizar Sep 25 '15 at 13:17
1
$\begingroup$

Alright so here is one way of going about it (if I understood you correctly). So let $Z = (z_1,z_2)$ be the point lying on the line through $A,B$ such that the line passing through $C,Z$ is perpendicular to the line through $A,B$. You can find this point as follows.

We can find that $d(A,C) = \sqrt{104}$, $d(C,D) = \sqrt{(z_1-12)^2 + (z_2-10)^2}$, and $d(A,D) = \sqrt{(z_1-10)^2 + (z_2-20)^2}$.

So Pythagorean's theorem gives us the equation: $$(d(C,D))^2 + (d(A,D))^2 = 104.$$

But we also know that $(z_1,z_2)$ is a solution to the equation of the line through $A,B$, namely the line $y= -\frac{13}{25}x + \frac{630}{25}$.

So now you have two equations and two unknowns. Give that a try and see if you can find $Z$. Once you have $Z$ you should be able to figure out whatever else you were wondering about.

$\endgroup$
  • $\begingroup$ Please see my edit. I added a picture showing what I need. Thanks. $\endgroup$ – Alireza Noori Sep 25 '15 at 14:26
  • $\begingroup$ @AlirezaNoori Alright well the same idea works to find $A'$. You don't need to find both $A'$ and $B'$ since you know $C$. So using Pythagorean's theorem you can find one equation in terms of the coordinates of $A' = (a_1,a_2)$. Then using point slope form, you can find the equation for the line through $A'$ and $C$ in terms of $a_1,a_2$. Again, two equations, two unknowns. $\endgroup$ – Ebearr Sep 25 '15 at 14:39
  • $\begingroup$ Thanks again. I'm really not that great at math. So could you give me the final formula or an example? I'm trying to create a software and it's urgent. I'd appreciate your help $\endgroup$ – Alireza Noori Sep 25 '15 at 14:54
  • $\begingroup$ I solved this with another method (you can see my solution at the end of my question) but I choose this as the answer. Thanks. $\endgroup$ – Alireza Noori Sep 28 '15 at 23:18
0
$\begingroup$

Maybe it can be of help knowing that the line you want passes through the midpoint of $AC$. So I think the simplest way to solve your problem is finding the midpoint $M=(A+C)/2$ and then drawing the line through $M$ having the same slope as $AB$.

$\endgroup$

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.