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Find the Coordinates of Point p

Date: 10/22/2002 at 03:09:45
From: Erwin Schwarz
Subject: Solving Radical Equations

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
Subject: Re: Solving Radical Equations

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. 

I hope this helps.  Write back if you'd like to talk more about this, 
or anything else. 

- Doctor Ian, The Math Forum 

Date: 10/22/2002 at 20:31:42
From: Erwin Schwarz
Subject: Solving Radical Equations

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
Subject: Re: Solving Radical Equations

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 

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 
Associated Topics:
High School Coordinate Plane Geometry

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