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In the Name of God بسم الله

Math Help

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Can anyone here help me real quick with this math question I have from Calc dealing with finding the closest point on a line to a given point?

 

Question: Find the point on the line y=4x+1 that is closest to the point (2,5)

My work so far http://i.imgur.com/OibGrWE.jpg

 

I honestly don't know if that is correct so far or not. I am not sure what to do with that y=29/5

Edited by Repentant
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Can anyone here help me real quick with this math question I have from Calc dealing with finding the closest point on a line to a given point?

 

Question: Find the point on the line y=4x+1 that is closest to the point (2,5)

My work so far http://i.imgur.com/OibGrWE.jpg

 

I honestly don't know if that is correct so far or not. I am not sure what to do with that y=29/5

It is very easy to find information about this using Google search.

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First Keep in mind that u have let D^2 = f(y)

That is very risky. Because it will give u the correct answer only if the minimum distance is greater or equal to 1. 

It is safer to let D = f(y)

 

There is also a mistake in your Fourth Line.

F'(y) = 2(y/4-9/4)+2(y-5)

 

it should be F'(y) = 2(y/4-9/4)*1/4 + 2(y-5)  because u also differentiate y/4 and times it in.

Doing that gives y=89/17

 

Now that u have y, sub it into the equation of the line to get ur x.

Then u have ur x and y coordinates for the point u r after. 

Edited by SlaveOfAllah14
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Its been ages since I looked at a problem like this. Do you find a point of tangency of a line with slope of 4 of a circle centered at (2,5)?

Sorry I'd have to pull out my books and review to help more, and my books are packed for moving. Hope you find your solution.

Edit: never mind. See sample problem above.

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I'm a bit rusty, but all I can say is that the point (2,5) has to be perpendicular to the line given.

 

Therefore the gradient of that perpendicular line will be the inverse of the gradient of the given line.

 

So when you get the equation of this perpendicular line, solve for x and y simultaneously and that will be your coordinates.

 

I didn't read everyone else's responses but hope this helps.

 

All the best.

Edited by JSAli
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There is a mistake in your Fourth Line.

F'(y) = 2(y/4-9/4)+2(y-5)

 

it should be F'(y) = 2(y/4-9/4)*1/4 + 2(y-5)  because u also differentiate y/4 and times it in.

Doing that gives y=89/17

 

Now that u have y, sub it into the equation of the line to get ur x.

Then u have ur x and y coordinates for the point u r after. 

 

What would I do with that y?

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What would I do with that y?

Once u get your y, u sub it into the equation of the straight line to find your x. But first sub your y into your F''(y) to show that at that y, F(y) is a minimum. Again it is safer that u let F(y) = D not D^2. That means u need to do ur differentiation again. 

Edited by SlaveOfAllah14
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It's a really simple question once you realize that it can be solved much easier using analytic geometry.

 

The closet point to another point on a straight line and not on a curve can be achieved by drawing a perpendicular line from that point toward the line.

 

Here your line is y=4x+1

All straight lines that are perpendicular to this line are in the form y=(-1/4)*x + C where C is the constant. (-1/4) was achieved by realizing the tangent relation of two perpendicular lines. Now we know one point from this line and that's (2,5). Substituting it into the equation yields C= 5.5. Now we have to find a point that exists on both these lines. That means solving the two equation, two unknown parameters system of y=4x+1 and y=(-1/4)*x + 5.5. That yields (89/17, 18/17)

 

Of course if the question had a curve in it instead of a line it would have been a totally different story.

Edited by Abu Nasr
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