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Find the point on the line
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is this for linear algebra?
nope
for calc
ok, so there are 2 main ways of doing this; with the orthogonal projection and the dot product
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oh and also with calculus of course
oh really? doesn't sound familiar
never mind lol forgot the easiest one was the cal one
oh ok lol
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.
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im sorry but im still lost. do i take the derivative of that then?
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