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yes, but its complicated :)

could you show me? thx alot!

since a square has diags that are perp to each other; we need 2 lines that meet in the middle of AC

C (8,4)
-A (0,2)
-------
8,2; stack y/x: 2/8 = 1/4 for our slope

perp slopes have the habit of being the negative reciprocal of each other. m1*m2 = -1

perp slope for th eline from B to D is then: -4/1

|dw:1327673742761:dw|

this is what weve got drawn up in out heads, or on paper

i know, this is where I had a brillant notion last time ... and I just remembered it :)

think of a pinwheel ....

|dw:1327674432861:dw|

our x and y parts for A to E are going to be swapped and negated for the perp from B to E

|dw:1327674563395:dw|

E (4,3) E (4, 3)
+(-1,4) + (1,-4)
------ --------
B (3,7) D (5,-1)

that should clean up the right path a little better :)

i kind of get it from the pinwheel ting

ok thx:)

the pinwheel thing is a very good visual to guide us :)