Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
@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
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.