# Interesting Proof

1. Mar 8, 2006

### ohlhauc1

We had to do a question in my Advanced Mathematics class, and the way that I did the problem was supposedly right. My teacher even did it that way. However, the answer was wrong, so my teacher showed us the correct way to do the problem.

The dilemma I have is that BOTH ways should work, but they don't. Therefore, I am going to give both solutions to the problem, and I was wondering if you could tell me why the first one does not work.

Question: The coordinates of points P and Q are (1,2) and (2,-3), respectively, and R is a point on the line x=-1. Find the coordinates of R so that PR + RQ is a minimum?

Here is the graph/image I used for this problem:
http://s62.yousendit.com/d.aspx?id=0...33IOBSY2IH1Z3A

MY SOLUTION:

Equations: PR^2 = 2^2 + y^2
QR^2 = 3^2 + (5 - y)^2

PR^2 + QR^2 = minimum or m
2^2 + y^2 + 3^2 + (5 -y)(5 - y) = minimum
4 + y^2 + 9 + 25 - 10y + y^2 = minimum
2y^2 - 10y + 38 = minimum

y = -b/2a = 10/4 = 5/2

Y-coordinate of R = 2 - 5/2 = -1/2

Therefore, the coordinates of R are (-1, -1/2)

CORRECT SOLUTION:

Let m = minimum

mP'R = [ (2 - y) / (-3 + 1) ] = [ (2 - y) / -2 ]

mRQ = [ (y + 3) / (-1 -2) ] = [ (y + 3) / -3 ]

Note: mP'R = mRQ
[ (2 - y) / -2 ] = [ (y + 3) / -3 ]
= -2y - 6 = 3y - 6
= -5y = 0
y = 0

The coordinates of R are therefore (-1, 0).

*So...I have presented both solutions, and I would highly appreciate your help. I am really interested in trying to understand this dilemma.

Thanks

2. Mar 8, 2006

### 0rthodontist

Your link doesn't work and I haven't gone through all your equations, but I notice this:
Minimizing the squares of each of the two distances is not the same as minimizing the sum of the two distances. The idea that is confusing you is the fact that minimizing the square of a distance is equivalent to minimizing the distance itself. However, the square of (PR + QR) is PR^2 + 2PR QR + QR^2, not just PR^2 + QR^2.

Last edited: Mar 8, 2006
3. Mar 8, 2006

### ohlhauc1

Thanks for the comment.

Try this URL:

http://spaces.msn.com/cloe-scibus/

The photo should be the first thing on the top and it is under PROOF PICTURE.

4. Mar 8, 2006

### AKG

I don't know if you understand the correct solution. Under CORRECT SOLUTION, you've written "Let m = minimum". That's not what's being done. P' is the point you get by relflecting P' in the line x=-1. Now mP'R denotes the slope of the line from P' to R, and mRQ denotes the slope of the line from R to Q.

When you reflect P in the line x=-1, the shortest distance from P' to Q is a straight line. Now if you find the point on x=-1 that the line P'Q crosses, then that will be the point R. Do you see why? So in order to find the point R, you need the slope of P'R to be the same as RQ, and that's what mP'R and mRQ mean. It's like mAB is the slope of the segment AB.

Your solution, I'm sorry to say, is nowhere close. For one, as 0rthodontist mentioned, you need to minimize PR + RQ, not PR² + RQ². Also, your equations for PR² and QR² are wrong. Well, it seems you've moved the points down 2 units. If you solve for y this way, then to solve the original problem, you will have to then add 2 to your answer. Instead, you subtract your answer from 2, which is wrong.

5. Mar 8, 2006

### ohlhauc1

I have verified my proof. Here it is:

Remember from the question that the x-coordinate is -1.

Note: PR^2 + QR^2 is not equal to PR + QR
PR^2 + 2PRQR + QR^2 = PR + QR = m

PR^2 = 2^2 + y^2
QR^2 = 3^2 + (5 - y)^2
PR = 2 + y
QR = 8 - y

PR^2 + 2PRQR + QR^2 = m
(2^2 + y^2) + [ 2(2 + y)(8 - y) ] + [ 3^2 + (5 - y)^2 ] = m
4 + y^2 + 2 (16 + 6y - y^2) + (36 - 10y + y^2) = m
4 + y^2 + 32 + 12y - 2y^2 + 36 - 10y + y^2 = m
2y + 70 = m

y = -b/2a = -2/2(0) = 0
y = 0

The coordinates of R so that PR + QR is a minimum are (-1,0).

6. Mar 8, 2006

### AKG

PR^2 + 2PRQR + QR^2 = PR + QR

This is wrong. Perhaps you mean:

PR^2 + 2PRQR + QR^2 = (PR + QR)²

Where are you getting these equations from:

PR = 2 + y
QR = 8 - y

The equation:

PR^2 = 2^2 + y^2

is correct. Now if somehow you know before hand that PR = 2 + y, you can solve for y = 0 immediately:

PR² = PR²
(2 + y)² = 2² + y²
2² + 4y + y² = 2² + y²
4y = 0
y = 0

However, I don't see how you could have gotten that equation for PR to start with. And your equation for QR is definitely wrong, since you have:

QR^2 = 3^2 + (5 - y)^2
QR = 8 - y

So we should get:

QR² = QR²
3² + (5-y)² = (8 - y)²
34 - 10y = 64 - 16y
6y = 30
y = 5

which is wrong. This is also wrong:

(2^2 + y^2) + [ 2(2 + y)(8 - y) ] + [ 3^2 + (5 - y)^2 ] = m
4 + y^2 + 2 (16 + 6y - y^2) + (36 - 10y + y^2) = m

3² + 5² is not equal to 36. This is also wrong:

4 + y^2 + 32 + 12y - 2y^2 + 36 - 10y + y^2 = m
2y + 70 = m

4 + 32 + 36 is not 70, it's 72. Curiously enough, had you computed 3² + 5² correctly, you would have gotten 34, so you'd be adding 4 + 32 + 34, which does happen to be 70. Finally, the following is wrong:

2y + 70 = m

y = -b/2a = -2/2(0) = 0

No, the first line isn't a quadratic expression. It has to take the form ay² + by + c, in which case the minimum is -b/2a. If you want to regard 2y + 70 in this form, then a = 0, b = 2, so you get -2/0, which is undefined, not 0. Think about it, 2y + 70 is a linear expression, it doesn't have a minimum. It might be better for you to think of the equation y = 2x + 70. Graph it, you know it has no minimum. It's just a line, it keeps going in opposite directions forever.