another word problem

Obsession

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Mar 4, 2006
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Marsha and her study partner, John, enjoy problem solving. They even challenge each other on occasion to solve problems that they themselves create.
One day, Marsha shows her friend a new problem she has just prepared, saying, "I'll bet you can't solve this little beauty."

John responds by singing a line from a famous Broadway show tune, "Anything you can do, I can do better!" He takes the paper she offers and reads these words.


A point (c,d) lies in the first quadrant. A line passing through (c,d) hits the positive y axis at P and the positive x axis at Q. If point O = (0,0), find the area of triangle POQ in terms of "c" and "d".
After drawing a simple sketch, John begins to frown. He looks worried that maybe he can't solve it after all. He says, "Gee, there seems to be something missing here."

Marsha glances at the paper and says, "Oh, you're so right. That was my first draft. There I had left out the important fact that triangle POQ is an isosceles one. I have that on this other paper."

With that new fact to use, John promptly provides an answer that satisfies Marsha. Write what John might have written for his solution.
----
Ok, having quite a few problems with this. I figured that we must use a graph, and the problem involves an isosceles triangle. and in POQ 'O' is the vertex(i hope....?)
but other than that...I'm totally lost.
:cry:
 
Hello, Obsession!

That's a really long story . . . most of it unnecessary.

A point (c,d) lies in the first quadrant.
A line passing through (c,d) hits the positive y axis at P and the positive x axis at Q.
If point O = (0,0) and triangle POQ is isosceles, find the area of triangle POQ in terms of "c" and "d".
If the triangle is isoceles, it must look like this:
Code:
      |
     P*
      | \
      |   *(c,d)
      |     \
      |       \
      |         \
    - + - - - - - * - -
      |O          Q

The two intercepts are equal and the slope of the line is -1.

The equation of the line is: \(\displaystyle \.x\,+\,y\:=\:p\)

Since \(\displaystyle (c,d)\) is on the line: \(\displaystyle \,c\,+\,d\:=\:p\)

And the intercepts are: \(\displaystyle \,(c+d,\,0)\) and \(\displaystyle (0,c+d)\)

\(\displaystyle \;\;\)Hence: \(\displaystyle OP\:=\:OQ\:=\:c+d\)

Therefore, the area of the triangle is: \(\displaystyle \,\frac{1}{2}(c+d)^2\)
 
i'm still confused, i get what you're explaining (well atleast a bit)...but why would the slope be -1?

----
edit:wait nevermind, i get it now
 
Code:
      |
     P*
      | \
      |   *(c,d)
      |     \
      |       \
      |         \
    - + - - - - - * - -
      |O          Q
Try it this way if you're confuded; using Soroban's Picassorial design:

draw perpendiculars from (c,d) to OP and OQ;
this will create 2 similar isosceles triangles inside triangle POQ;
then you'll easily see that OP = OQ = c + d.
 
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