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Intersection Points


Date: 03/13/99 at 04:55:17
From: sara
Subject: Intersection Points

A line is drawn through the point A(1, 2) to cut the line 2y = 3x - 5 
in P and the line x + y = 12 in Q. If AQ = 2AP, find the coordinates 
of P and Q.

The answers given in the book are P(4, 3.5)  Q(7, 5) or P(2/5, -19/10) 
Q(2 1/5, 9 4/5)

I have no idea how they got those answers.

Please help.


Date: 03/13/99 at 07:34:04
From: Doctor Mitteldorf
Subject: Re: Intersection Points

Here is how I would tackle it if I were you.  First, I would get a 
piece of graph paper and find the point A, plot the two lines, and 
draw some lines from A out in the direction of those two lines to see 
how the line can be situated to make AP about half the length of AQ.  
So you have solved the problem approximately 'by eye'. That will give 
you a feel for what you are being asked to do.

The next thing I would do would be to try the answers that the book 
gives you: convince yourself that point P is really on line 
2y = 3x - 5. Find the distance between the point P and A. Do the same 
for Q, and see that their solution works. Do the same for their second 
solution.  
----------------
Next, I would say: suppose the slope of the line we draw from A is m.  
Then, we can write the equation of the line as (y - 2)/(x - 1) = m. 
Solve this equation simultaneously with the first equation 2y = 3x - 5 
and you will have an expression for the coordinates of P in terms of 
m. Now subtract the coordinates (1, 2) of the point A, and use the 
Pythagorean theorem to find an expression for the distance between 
point P and point A, all in terms of your one variable, m.

Now do the same thing for point Q: end by finding an expression for 
the distance between A and Q, written as an expression in m. Then you 
are finally ready to write an equation in m that says one expression 
is twice the other. If you are lucky, this will not be too messy, and 
you will be able to turn it into a quadratic equation for m. 

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


Date: 07/19/99 at 19:36:45
From: Tim Greene
Subject: Re: Intersection Points

Dr. Math -

My first inclination for solving this problem was apparently the 
solution you described (although I did not follow through on the 
details) - to solve the problem by using the Pythagorean theorem 
(distance formula) on segments AP and AQ and the fact that the slopes 
of segments AP and AQ are equal. When I attacked the problem that way, 
the algebra quickly got ugly, so I digressed to look for a quicker and 
easier approach.

I had first, as you suggested at the beginning of your solution 
description, sketched a rough graph of the problem, and by examination 
I had determined that there would be two solutions: one line through A 
satisfying the conditions of the problem will have a small slope, with 
points P and Q in the same direction from A; another line through A 
satisfying the conditions of the problem will have a much steeper slope, 
with points P and Q on opposite sides of A.

I began by labeling the two sets of solution points as P1 = (a,1.5a-2.5) 
and Q1 = (b,12-b); and P2 = (c,1.5c-2.5) and Q2 = (d,12-d). My 
simplification of the problem came from using the distance formula not 
on the diagonal segments AP and AQ but on the x- and y-components of those 
segments. For points P1 and Q1, for example, the equations which say that 
the x- and y-components of AQ are twice the corresponding components of AP 
are (b-1) = 2(a-1) and (10-b) = 2(1.5a-4.5); this is a set of linear 
equations which quickly gives the solution P=(4,3.5) and Q = (7.5). 
Similarly, the same approach for points P2 and Q2 gives another set of 
linear equations which give the solution P = (0.4,-1.9) and Q = (2.2,9.8)

- Tim Greene
    
Associated Topics:
High School Basic Algebra
High School Coordinate Plane Geometry
High School Geometry

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