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Mathematics 8 Online
OpenStudy (anonymous):

Find the point on the line

OpenStudy (anonymous):

OpenStudy (p0sitr0n):

is this for linear algebra?

OpenStudy (anonymous):

nope

OpenStudy (anonymous):

for calc

OpenStudy (p0sitr0n):

ok, so there are 2 main ways of doing this; with the orthogonal projection and the dot product

OpenStudy (p0sitr0n):

oh and also with calculus of course

OpenStudy (anonymous):

oh really? doesn't sound familiar

OpenStudy (p0sitr0n):

never mind lol forgot the easiest one was the cal one

OpenStudy (anonymous):

oh ok lol

OpenStudy (p0sitr0n):

so you have the distance \[D=\sqrt(6(x+2)^2+3(y-5)^2)\] this is the polynomial that describes the distance. Now what you can do is just take the derivatives to x and to y and then solve them for 0. As you know , P is an extremum if and only if the derivative f'(P)=0.

OpenStudy (anonymous):

im sorry but im still lost. do i take the derivative of that then?

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