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Find the point on the line $y = x + 2$ that is nearest to the point $(1,1)$. The shortest distance from point to point.

I honestly don't even know where to begin with this one.

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@AlexBecker: Given that this question is tagged calculus and the question to which you linked does not have a calculus-based answer (only geometric answers), I'm not sure that's a good choice of question to close this as a duplicate of. –  Isaac May 4 '12 at 0:11
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1 Answer 1

up vote 3 down vote accepted

Hint: If you want to use calculus, let $x$ be the horizontal coordinate of the point on the line. Then the point is $(x,x+2)$. You can calculate the distance from this to $(1,1)$ as a function of $x$, set the derivative to $0$.

Alternately, the shortest distance is along a perpendicular. Do you know the relation between the slope of a line and the slope of the perpendicular? Make a line through $(1,1)$ with that slope and find the intersection with your line.

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I'm not sure I understand, how would I create a function out of that? I'm trying to picture it, but I think I'm overlooking something really simple. –  Math_Phase May 8 '12 at 23:25
@Math_Phase: The distance from $(1,1)$ to an arbitrary point on the line is $d=\sqrt{(1-x)^2+(1-(x+2))^2}$. You can take $\frac{dd}{dx}$ and set it to $0$. –  Ross Millikan May 8 '12 at 23:30
Ah, I overlooked using the distance formula. Thank you –  Math_Phase May 8 '12 at 23:37
@Math_Phase Also note that often it is easier (i.e. less formulas to write) to calculate the point where $d^2$ attend it's minimum. –  AD. May 13 '12 at 5:59
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