Please a little help!?
823237
There is several way to solve the problem, but I think it is easies, to find the parametric representation, and then use, that the scalar-product of the orientations-vectors is zero, if the line are perpendicular, and the orientations-vectors is linear depended if parallel.\n\nL1:  y = -½x-3 => (x,y)= (1,-½)*t+(0,-3)\n4x-2y+5 = 0 <=> y=2x+5/2=>\nL2: (x,y) = (1,2)*t + (0+5/2)\n\nbecause (1,-½)*(1,2)=1-1=0 the lines are perpendicular
986490
Dude! Im having the same problems! Somebody HELP US!!!
886819
I might help you if I knew what you were exactly talking about.\nA little more info please.\n\nTHX
298754
Great answers so far!  I ADDED A LINK BELOW, that I think speaks to your question.\n\nLet's take this question to a vote, to get this resolved.\n\nWhat do you say?
278724
Please be more specific!!
286129
Maybe I can help.\n\nmahc_u00@yahoo.com
387050
welp cant help u there! thanks 4 the *2pts*!
438610
plz take help from yahoo. \ni don't know
609289
I wish I could help, but I have no earthly idea.\n\nI'm sort of glad that you asked the question here though, rather than in person ! ;-P
259900
Maybe a little bit! ;0)
825335
this might help...   ?
903963
No, I cannot help you.\n\nI can only help myself.
694063
Is anyone has a? Does anyone could help?
666601
Try this person, very clever!\nhttp://answers.yahoo.com/my/message_do?kid=AA10894849
1075916
I don't know what to do sorry!
732086
DON'T TRY TO FIGURE OUT ANYTHING. DON'T WORRY ABOUT IT. FORGET THE SO CALLED WANNA  BE FRIENDS TOO. THIS SOUNDS LIKE A BUNCH OF KID GAMES TELL THEM YOU DON'T HAVE TIME FOR IT. PICK YOUR FRIENDS LIKE YOU WOULD PICK A APPLE FROM A TREE. EVIDENTLY YOU HAVE SOMETHING THAT WOULD MAKE THEM HATE YOU ANYWAY.SO HOLD YOUR HEAD UP HIGH, SMILE, AND LAUGH THAT DUMB SHIT OFF
935888
????????????????????\n\nHelp please?