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### Find the Coordinates of Point p

```Date: 10/22/2002 at 03:09:45
From: Erwin Schwarz

I'm trying to help my daughter with her Grade 11 math. The chapter is
titled Solving Radical Equations and Inequalities, but I'm struggling
with helping her on a particular question that goes like this:

Point P lies on the line y=x. The coordinates of Q and T are (-1,8)
and (8,1) respectively. The sum of the distances QP and TP is 18
units. Determine the coordinates of P.
```

```
Date: 10/22/2002 at 12:07:39
From: Doctor Ian

Hi Erwin,

A picture usually helps:

Q  |              P         Q = (-1,8)
|          .
|       .
|    .
| .               T      T = (8,1)
---------- --------------------

If the coordinates of P are (p,p) (it's on the line y=x, so the
coordinates must have equal values), then

QT = sqrt[(p - -1)^2 + (p - 8)^2]

PT = sqrt[(p - 8)^2 + (p - 1)^2]

(I just used the standard distance formula,

distance from (x,y) to (x',y') = sqrt[(x - x')^2 + (y - y')^2]

which can be derived from the Pythagorean Theorem.)

And you're told that

QT + PT = 18

The only thing that makes the problem workable is that you just have
the one value, p, for both coordinates of P, since that gives you an
equation in one variable.

or anything else.

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 10/22/2002 at 20:31:42
From: Erwin Schwarz

Yes, I see now, and to carry it further via substitution it becomes
sqrt[(p-1)^2+(p-8)^2] + sqrt[(p-8)^2+(p-1)^2] = 18 and then you
remove the radical by squaring both sides of the equation.  Doesn't
that leave the right side of the equation really messy, or am I
missing some other form of reduction?

P.S. Thanks for your help - this is an awesome webpage
```

```
Date: 10/22/2002 at 22:37:50
From: Doctor Ian

Hi Erwin,

Yes, it's going to be a mess.  I haven't worked it through to the
end, but I'm guessing that you'll have to square the equation twice.
It's easier to see why if we use simpler expressions:

__     __
c = \| a + \| b

__   __
c^2 = a + 2 \| a \| b  + b

___
c^2 - a - b = 2\| ab

(c^2 - a - b)^2 = 4ab

Good luck!

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 10/23/2002 at 03:05:25
From: Erwin Schwarz
Subject: Thank you (Solving Radical Equations)

Excellent, I believe I can explain this to my daughter in such a way
she will understand.  Once again, this is an awesome service. Thank
you.
```
Associated Topics:
High School Coordinate Plane Geometry

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