# Stumped by a problem that would have been easy for me in high school.

• Jun 11th 2014, 01:54 PM
KeeganT
Stumped by a problem that would have been easy for me in high school.
Yep. My skills are far deteriorated, so much so that the problem I'm about to describe I'd have cringed at needed help with way back when. Anyways.....

Suppose I have a rectangle, $w wide and$h tall. Consider the diagonals going from top-left to bottom-right and from bottom-left to top-right. I will be cycling through each pixel, and I want to figure out for each one what is the distance from that pixel to the nearer diagonal.

I've tried to work it out on paper but ended up getting frustrated twice.

Can anyone help?

THX
KeeganT
• Jun 11th 2014, 08:01 PM
Prove It
Re: Stumped by a problem that would have been easy for me in high school.
I've noticed you have marked this as solved. Would you like to post your solution? Someone who has a similar problem might find it useful...
• Jun 12th 2014, 03:20 AM
KeeganT
Oops, didn't mean to mark it Solved.
Yeah, I'm still looking for a solution.

KeeganT
• Jun 12th 2014, 03:26 AM
topsquark
Re: Stumped by a problem that would have been easy for me in high school.
FYI: I have removed the "Solved" tag.

-Dan
• Jun 12th 2014, 08:41 AM
Soroban
Re: Stumped by a problem that would have been easy for me in high school.
Hello, KeeganT!

Quote:

Suppose I have a rectangle, $w$ wide and $h$ tall.
Consider the diagonals going from top-left to bottom-right and from bottom-left to top-right.
I will be cycling through each pixel, and I want to figure out for each one
what is the distance from that pixel to the nearer diagonal.

I placed the rectangle on a coordinate system.
Code:

  (-w/2,h/2)      |        (w/2,h/2)       B * - - - - - + - - - - - * A         |  .      |      .  |         |      .  |  .      |     - - + - - - - - + - - - - - + - -         |      .  |  .      |         |  .      |      .  |       C * - - - - - + - - - - - * D   (-w/2,-h/2)      |      (w/2,-h/2)                     |
Diagonal $AC$ has the equation: $\; y = \tfrac{h}{w}x \quad\Rightarrow\quad hx - wy \:=\:0$

Diagonal $BD$ has the equation: $\;y = \text{-}\tfrac{h}{w}x \quad\Rightarrow\quad hx + wy \:=\:0$

If the point $P(x_1,y_1)$ is in Quadrant 1 or 3, it is closer to $AC.$
The distance is: $\;d \:=\:\dfrac{|hx_1 - wy_1|}{\sqrt{h^2+w^2}}$

If the point $P(x_1,y_1)$ is in Quadrant 2 or 4, it is closer to $BD.$
The distance is: $\:d \:=\:\dfrac{|hx_1+wy_1|}{\sqrt{h^2+w^2}}$

I hope this helps.