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Mathematics 19 Online
OpenStudy (anonymous):

Find the coordinates of midpoint P of BC. Then calculate lengths AP, BP, and CP.

OpenStudy (anonymous):

OpenStudy (anonymous):

you find mid poiint exactly right..now to find AP BP CP use distance formula...as d=sqrt{(x2-x1)^2+(y2-y1)^2}

OpenStudy (anonymous):

So, AP=sqrt{(h-0)^2+(k-0)^2}?

OpenStudy (anonymous):

Would AP=hk?

OpenStudy (anonymous):

What do u mean by that?

OpenStudy (anonymous):

@zaibali.qasmi

OpenStudy (anonymous):

AP=sqrt{h^2+k^2}

OpenStudy (anonymous):

next../

OpenStudy (anonymous):

nice baby...i think your daughter..??so cute bothf you///

OpenStudy (anonymous):

So, it's not AP=h,k?

OpenStudy (anonymous):

no..AP mean you have to find that distance..its not coordinates..its magnitued..!!

OpenStudy (anonymous):

I'm a little confused. What would the answer be?

OpenStudy (anonymous):

AP=sqrt{(h-0)^2+(k-0)^2} => AP=sqrt{h^2+k^2} its answer.

OpenStudy (anonymous):

So then, BP=sqrt{{h-0)^2+(k-2k)^2} => BP=sqrt{h^2+-k^2}

OpenStudy (anonymous):

yes..but (-k)^2=k^2 so it will become same as that of AP...

OpenStudy (anonymous):

Thanks, you were a lot of help!

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