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Finding Three Lines that Form a Triangle


Date: 08/30/2001 at 16:30:56
From: Joe Kowalski
Subject: Triangle Area Puzzle

The question is: 

Find an equation for each of the lines that goes through the point 
(3,2) forming a triangle of area 12 with the coordinate axes. There 
are three answers.

I have been able to come up with one answer, which is y = (-2\3)x+4. 
The only way I've been trying so far is trial and error. My teacher 
said we really wouldn't be able to figure it out any other way with 
our current knowledge.


Date: 08/31/2001 at 09:25:14
From: Doctor Rob
Subject: Re: Triangle Area Puzzle

Thanks for writing to Ask Dr. Math, Joe.

In general, there can be as many as four such triangles, but in the
particular case you were given, the number is just three.  Here is
a rough diagram:

                           ^y
                           |
                           |
                         A o.
                           | `-.
                           |    `-.
                           |       `-.
                           |          `-. P(3,2)
                           |           __`o.
                         C |   __,,--''_- : `-.
                       __,,o-''      _-   :    `-.
               __,,--''    |       _-     :       `-.      x
   --------o---------------o------o-------o-----------o--->
           D              O|   _- E       Q           B
                           | _-
                         F o
                           |
                           |

(Actually, points D and F are much farther from the origin than
shown in the diagram.)

The three triangles are OAB, OCD, and OEF. You have found OAB by
finding line AB to be y = (-2/3)*x + 4, and lengths OA = 4 and
OB = 6.

If you know something about similar triangles, the formula for area, 
how to solve quadratic equations, and how to calculate with 
expressions involving square roots, you can find all three lines
exactly, and without trial-and-error.

To find CD, drop a perpendicular from P to the x-axis, meeting it at 
point Q. We know length QP = 2 and OQ = 3. Then the right triangles 
DPQ and DCO are similar, because they share angle D and each has a 
right angle (<DQP and <DOC), so all three angles must be respectively 
equal. Call length DO a, and length OC b. Then we have two equations 
involving a and b: one from the area being equal to 12, and the other 
because corresponding sides of similar triangles are proportional:

   12 = area = (1/2)*a*b,
   b/2 = OC/QP = DO/DQ = a/(a+3).

From the first equation, b = 24/a, and, substituting that into the
second equation, we get

   12/a = a/(a+3),
   12*(a+3) = a^2,
   0 = a^2 - 12*a - 36,
   a = 6 + 6*sqrt(2)  (ignoring the negative root as extraneous),
   b = 24/a,
     = 24/(6+6*sqrt[2]),
     = 4/(1+sqrt[2]).

Now to rationalize the denominator, multiply both numerator and
denominator by -1 + sqrt(2):

   b = 4*(-1+sqrt[2])/([1+sqrt(2)]*[-1+sqrt(2)]),
     = 4*(-1+sqrt[2])/(-1+2),
     = -4 + 4*sqrt(2).

Now the points (0,b) and (3,2) are on the line (so is (-a,0), but we 
only need two), so we can find the equation of line CDP using the 
two-point form:

   (y-y1)*(x2-x1) = (y2-y1)*(x-x1),
   (y-b)*(3-0) = (2-b)*(x-0),
   3*(y-b) = (2-b)*x,
   y = ([2-b]/3)*x + b,
   y = ([6-4*sqrt(2)]/3)*x + 4*(-1+sqrt[2]).

The analysis of the other case, line EFP, is similar, using the fact 
that right triangles EQP and OEF are similar. We get that the equation 
of line EFP is

   y = ([6+4*sqrt(2)]/3)*x + 4*(-1-sqrt[2]).

Do you think this is beyond what you know?

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Equations, Graphs, Translations

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