find point(x,y) on the line y=x using points (-8,10) and(8,-7)
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ohhh sorry i for got to ADD equidistant point lol
:) you want the point that is equidistant from the givens and that sits on y=x then right?
yes lol :)
in other words; the place tha thte line cross
what kind of a line can we create with the 2 given points?
or am i thinking of this erroneously .....
we need a distance formula perhaps
welll yess i know we need the distance formula however since were lookimg for a set of points what would we replace it by in the equation
(-8,10) (8,-7) -(x , y) -( x, y) ----------------------- -8-x,10-y 8-x,-7-y these are our respective distances from any point on y=x
this always gets messy when I try to do it :)
LOL :P at least youre honest. i started doing it and threw out half my copybook
\[\sqrt{x^2+16x+64+y^2-20y+100}\] equals \[\sqrt{x^2-16x+64+y^2+14y+49}\] as long as I did the squareing right that looks to be getting closer
we can square each one to get rid of the radicals and try to clean it up some: \[x^2+16x+y^2-20y+164=x^2-16x+y^2+14y+113\] \[16x-20y+164=-16x+14y+113\] \[32x-20y+164=14y+113\] \[32x+164=34y+113\] \[32x+51=34y\] hmmm
i kinda get it but for some reason im also a little confused lol
yeah, its trying to use a distance formula to equate the distance from one point to the y=x line; with the other point
i think where this line and y=x cross is the point of interest tho
lets try it out :) \[32x+51=34y\] and since y=x \[32x+51=34x\] \[51=2x\] \[51/2=x\] perchance?
http://www.wolframalpha.com/input/?i=%2851%2F2%29^2%2B16%2851%2F2%29%2B%2851%2F2%29^2-20%2851%2F2%29%2B164%3D%2851%2F2%29^2-16%2851%2F2%29%2B%2851%2F2%29^2%2B14%2851%2F2%29%2B113++ the wolf says im good
thats it! thats the answer to it i just wanted to know how to come up with that. such a simple problem turns out to be not so simple. i very much appreciate it:)
just tough out the distance formulas and equate them \[\sqrt{(x-Px_1)^2+(y-Py_1)^2}=\sqrt{(x-Px_2)^2+(y-Py_2)^2}\]
thanks youve been a great help. midterms are the worst! lol
youre welcome :) good luck
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