# Finding the coordinates and distance of this drawing

Hello everyone, I'm having problems figuring out the distance and the cordinates of the point P(s,t). The directions are in the drawing posted. I'm not sure if i drew the picture right though. I used the distance formula to find the distance between the origin i named <x,y> and the point P(s,t) which is on the vector a = <x1,y1>. But i might have screwed that up too because I assumed its the midpoint of the vector. I don't know how i'm suppose to simplfy |d| in the drawing. Once thats simplified i'm also confused on how that is going to tell me the coordinates of P. Thanks,

http://show.imagehosting.us/show/654150/0/nouser_654/T0_-1_654150.jpg

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DaveC426913
Gold Member
You didnt't attach it.

Sorry about that, thanks for the info, it is now attached!

Fermat
Homework Helper
P is a point on the vector A, so P's position vector is given as a fraction (or multiple) of the position vector of A. i.e. OP = λ(x1,y1).

If P were the mid-point of OA (which it isn't) then you would have λ = ½.

You can work out the slope of OA (=m, say) using the coords of A and you know that BP is perpindicular to OA, so you can write the slope of BP in terms of m. You can also work out the slope of BP using the coords of P and B. Now solve for λ, and you can then work out the coords of P and the length of OP.

thanks for the reply but i'm lost.
You can work out the slope of OA (=m, say) using the coords of A and you know that BP is perpindicular to OA, so you can write the slope of BP in terms of m. You can also work out the slope of BP using the coords of P and B. Now solve for λ, and you can then work out the coords of P and the length of OP.

Where is the point OA coming from? by OA do u mean the vector O standing for origin and A for the a vector? So your saying set BP, B is a vector <x2,y2> and P is a point with coords (s,t) = m. Am i allowed to do this? BP = (y2-t)/(x2-s); or are u not talking about m = (y2-y1)/(x2-x1)?