How to Find points for draw a Perpendicular line of given two points

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Hi,
I have two point (x1,y1) and (x2,y2) which makes a straight line. Now I want to draw two lines through the terminal points which will be perpendicular to the main line.

in attached image you can see what i need.

Pointsfind.png - Google Drive[^]

if anyone know how to do please help
Thanks in Advance.

Posted 29-Dec-15 20:09pm

MinhajAli

Updated 30-Dec-15 2:02am

v4

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Comments

Richard MacCutchan 30-Dec-15 5:03am

You would do it by writing some code. Your link does not seem to contain any useful information, please edit your question and explaion in better detail.

[no name] 30-Dec-15 9:13am

A vector prependicular to (a,b) is (-b,a). This one can prove by the help of scalar product.
In your case (a,b)= (x2,y2)-(x1,y1).
Now with this (-b,a) vector one can easily calculate the four (?,?) vectors respectively their representants.

Sergey Alexandrovich Kryukov 30-Dec-15 13:10pm

The problem is ambiguous, because you can draw multiple perpendicular lines.
—SA

[no name] 30-Dec-15 13:44pm

Nothing is ambiguous, except the abs() of the perpendicular vectors.

Sergey Alexandrovich Kryukov 30-Dec-15 15:46pm

You are right: I noticed that the inquirer needs two lines "through the terminal points". Of course, this is solvable. And "abs of perpendicular vector" also does not make things ambiguous, because the question is not about the vector. It's enough to point out two points per straight line, "representative enough"; ideally, the points at the interception of these lines to the viewport boundaries (or beyond this boundary).
—SA

2 solutions

Solution 1

A vector perpendicular to (a,b) is (-b,a). This one can prove by the help of scalar product (also named dot product).

In your case (a,b)= (x2,y2)-(x1,y1). Now, with this (-b,a) vector one can easily calculate the four (?,?) vectors respectively their “representants” (sorry I do not find the right word in English).

Because (?,?) is not specified more precise, I personally would use the above mentioned calculations. Another way is to use “unit vector” for (-b,a), but in case of graphics one has to pay attention, not to lose precisions.

Sorry for my bad English, I hope it helps.

It is simple vector computation. Even if others here think it is more.

Posted 30-Dec-15 6:36am

User 11060979

Updated 30-Dec-15 8:27am

v2

Comments

Sergey Alexandrovich Kryukov 30-Dec-15 12:52pm

What are you talking about?

How (a, b) and (a, -b) can be perpendicular? Can't you see that (100, 1) and (100, -1) are positioned at sharp angle?

"Find a perpendicular line" is ambiguous. The question is not about finding a perpendicular vector, it's about finding two points. You can draw multiple perpendicular lines...

—SA

[no name] 30-Dec-15 12:57pm

Sorry, I forgot to mention the trivial Solutions as an "exception"... but I think OP Needs first to consume the non trivial ones :-)

It is up to you to describe it more precisely.
Bruno

This comment is not longer valid after edit of SA's comment

Sergey Alexandrovich Kryukov 30-Dec-15 12:59pm

Sorry, I fixed my comment above. The whole problem is posed ambiguously, and your post is plain wrong.
Please read again.
—SA

[no name] 30-Dec-15 13:19pm

Please explain "and your post is plain wrong". Thank you.

Sergey Alexandrovich Kryukov 30-Dec-15 15:41pm

My mistake, sorry. I'll try to fix it.
—SA

[no name] 30-Dec-15 13:26pm

Please, (a,b)= (100,1). What is (-b,a)? I'm very sure it is (-1,100) and _not_ (100,-1)....
And scalar product, is it not valid anymore?

[no name] 30-Dec-15 13:34pm

Better think again about it, I think. I kindly ask you to read all again.
Better to remove your comment!
Bruno

You are usually very precise. Now again, what is about this:
<blockquote class="quote"><div class="op">Quote:</div>What are you talking about?

How (a, b) and (a, -b) can be perpendicular? Can't you see that (100, 1) and (100, -1) are positioned at sharp angle?

"Find a perpendicular line" is ambiguous. The question is not about finding a perpendicular vector, it's about finding two points. You can draw multiple perpendicular lines...

Better remove this post, or really solve this problem...

—SA</blockquote>

"How (a, b) and (a, -b)": Did I write this? No!!!!

Sergey Alexandrovich Kryukov 30-Dec-15 15:40pm

Sorry, you didn't.
—SA

[no name] 30-Dec-15 15:42pm

Yes, I did not. And all the rest of my answer makes a lot of sense!
Upset about the explicit uncommented downvote from you!
Bruno

Sergey Alexandrovich Kryukov 30-Dec-15 15:47pm

No problem, it was just my mistake which I'm fixing; no need to be upset. I always try to fix my mistakes is some prove me wrong.
Sorry, but don't be upset.
—SA

[no name] 30-Dec-15 16:06pm

Thank you! I just have some doubts about my knowledge of vector geometry.
Bruno

Sergey Alexandrovich Kryukov 30-Dec-15 16:10pm

Not at all.
But the question still needs more complete solution. This part of algebra, called "analytical geometry" (it's really algebra, no mistake here) is not too hard to learn.
—SA

[no name] 30-Dec-15 16:15pm

"But the question still needs more complete solution":
I'm keen on reading it from your Point of view :-). Bruno

[no name] 31-Dec-15 10:51am

Hard earned forty Points :-)
Bruno

MinhajAli 31-Dec-15 1:08am

Thanks all for your reply.
but still i need the solution.

[no name] 31-Dec-15 9:06am

According to your example:
(x2,y2)= (2,6)
(x1,y1)= (6,1)

(a,b)= (x2,y2) - (x1,y1)
(a,b)= (2,6) - (6,1)= (2-6, 6-1)= (-4,5)

Prependicular:
(-b,a)= (-5,-4) // This one you can scale to make the graphics nice

?2 = (x2,y2) - (-b,a)= (2,6) - (-5,-4)= (2- -5,6- -4)= (7,10)
?2'= (x2,y2) + (-b,a)= (2,6) + (-5,-4)= (2-5, 6-4) = (-3,2)

?1 = (x1,y1) - (-b,a)= ....
?1'= (x1,y1) + (-b,a)= ....

Finally draw the two lines from ?2 to ?2' and from ?1 to ?1'

I suggest to read trough something like this:Vectors[^]

MinhajAli 1-Jan-16 12:07pm

Thanks for your reply. Actually i am week in this area.
I am trying it. I will get back to you.

[no name] 1-Jan-16 12:30pm

You are very welcome!
I don't think that you are weak on this. You simply did not learn it until now. I had the big opportunity to learn all this stuff from a very good teacher. Don't give up please! Vector algebra looks may be difficult at the beginning, but it is much more easy than you expect. Go on with it, it helps a lot and makes things easy.

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MinhajAli · Accepted Answer · 2016-02-10T00:29:00

Thanks to all for your reply.
i did the same by using the following code

dx = x1-x2
dy = y1-y2
dist = sqrt(dx*dx + dy*dy)
dx /= dist
dy /= dist
x3 = x1 + (N/2)*dy // where N is the length of line to be draw.
y3 = y1 - (N/2)*dx
x4 = x1 - (N/2)*dy
y4 = y1 + (N/2)*dx

and same way i found the x5,y5 and x6 y6.