Jump to content

View more

Image of the Day

Boxes as reward for our ranking mode. ヾ(☆▽☆)
#indiedev #gamedev #gameart #screenshotsaturday https://t.co/ALF1InmM7K
IOTD | Top Screenshots

The latest, straight to your Inbox.

Subscribe to GameDev.net Direct to receive the latest updates and exclusive content.


Sign up now

the math challenge

4: Adsense

Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.


  • You cannot reply to this topic
4 replies to this topic

#1 Pseudo   Members   

100
Like
Likes
Like

Posted 21 July 2001 - 12:46 PM

I have 2 2d line equations in general form (Ax+By+C=0) and I would like to find the point of intersection (if any). How can I do this? I started to write it out on paper and gave up because I would have had to simplify a huge equation. I know there is a simple solution, so if you are man(or women) enough to take the math challenge, please post what you think is the quickest solution. Pseudo

#2 sympathy   Members   

122
Like
Likes
Like

Posted 21 July 2001 - 12:57 PM

Lets say we have two lines:

y = mx + b

y = 3x + 3
y = 4x + 1

To find the point of intersection we arrange them like this:

x + y = b

-3x + y = 3
-4x + y = 1

This is a systems of equation. What we have to do is have either the x or the y on the top and bottom being the inverse value. So lets use the x:

12x - 4y = -12
-12x + 3y = 3

All I did was multiply the top by - 3, and the bottom by + 3. This gives us an equation we can solve. We now cross of the zeroing xs, add the ys and add the ending term:

0x - 1y = -9
-1y = -9
y = 9

Now to find the x point, we take a line and substitute in for y and solve for x.

9 = 3x + 3
6 = 3x
2 = x

(2, 9) is the point of intersection for the two lines defined by: y = 3x + 3, y = 4x + 1. Hope this helped. I did this based on memory so I''m not sure if it is all entirely correct.



#3 Anonymous Poster_Anonymous Poster_*   Guests   

Likes

Posted 21 July 2001 - 01:00 PM




I''d rewrite each formula so that a single variable (x or y) was on one side. I will pick y:

Ax + By + C = 0
By = -Ax - C
y = -Ax/B - C/B (y=mx+b format)

Do that for each line, then set the y of one line = to the y of the other line, so you will get something like:

4x + 5 = 5x + 3

Then solve for x:

5 - 3 = 5x - 4x

2 = x








#4 Pseudo   Members   

100
Like
Likes
Like

Posted 21 July 2001 - 01:34 PM

thanks alot. I had a little brain fart I guess.

#5 Strife   Members   

374
Like
Likes
Like

Posted 21 July 2001 - 03:28 PM

Yeah, those are the two easiest ways to do it. If you have more than 2 equations, the same thing goes, only for n equations, you must have n variables to solve.


Got Slack?
Commander M




Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.