Please help me!! http://i.imgur.com/dlAh7BY.jpg?1
gotta use the slope somehow
suppose \(x\) was \(5\) for example how would you find \(y\) ?
i would find the slope using that point in the middle and the coordinate with a 5. then, i would make that slope equal to the slope of the given point and the point with the y. then i would solve for y. sorry if my explanation sucks. ;__; is that how we do it?
yes that is how i would do it i take it for granted that you know that the length, by pythagoras, is \[\sqrt{x^2+y^2}\] so our real job is to find \(y\) in terms of \(x\) and plug it in to that formula
lets forget the \(5\) and do it just with variables what is the slope of the line through \((0,y)\) and \((x,0)\) ?
oops did i lose you?
well, i think it's -2/ x-3 but is it also 2-y/3
then do i make it equal to each other and solve for y?
yeah you can do it that way too, i was going to make it easier
oh xD
actually i did it your way first, and we can do that if you like
but then i thought it would be easier if we only computed the slope between \((0,y)\) and \((x,0)\) which you get pretty much straight away as \(-\frac{y}{x}\)
then compute the slope between \((3,2)\) and \((x,0)\) and get \[\frac{2}{3-x}\] or as you had it \[\frac{-2}{x-3}\]
then yes, set them equal and get \[-\frac{y}{x}=-\frac{2}{x-3}\]
ohhh
well i got y = 2x/x+3 tho xD
same idea exactly only this one is easier to solve for \(y\)
i think you have as small mistake i think the \(+\) should be a \(-\) in the denominator
i.e. \[y=\frac{2x}{x-3}\]
ohhhh. i see what i did wrong. i accidently switched the sign at one point.. xD
and now we are done right? wasn't too bad
yea ur right. :D Thanks. but the final answer will be under a square root right?
yeah and squared too
okay ty :)
yw
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