1. Determine the co-ordinates

This is my first post. I graduated from mathematics in university 16 years ago so I am somewhat out of practice. I recently ran into my high school teacher from several years ago and he posed a math problem to me. I need some help with it, even a push in the right direction. Thanks. Here is the problem:

A line segment has its endpoints on the x and y axes.
The line segment passes through point (6,6).
The line segment's length is 21.
Determine the unknown points on the x and y axes.

Thanks
Norm F

2. Re: Determine the co-ordinates

Let the x-intercept be denoted by (a, 0)
Let the y-intercept be denoted by (0, b)
Then a^2 + b^2 = 21^2

Now draw the rectangle with vertices (0, 0), (6, 0), (0, 6), and (6, 6)
Draw a line segment from the y-axis to the x-axis through (6, 6)

There should be a triangle above the rectangle, and one to the right. These two will be similar, and using proportions (along with the pythagorean theorem, above) should give the solution.

3. Re: Determine the co-ordinates

Originally Posted by TheChaz
Let the x-intercept be denoted by (a, 0)
Let the y-intercept be denoted by (0, b)
Then a^2 + b^2 = 21^2

Now draw the rectangle with vertices (0, 0), (6, 0), (0, 6), and (6, 6)
Draw a line segment from the y-axis to the x-axis through (6, 6)

There should be a triangle above the rectangle, and one to the right. These two will be similar, and using proportions (along with the pythagorean theorem, above) should give the solution.
Thanks for your reply. What you've said, is what I have done. The rectangle (square) is 6 by 6, and the diagonal from (0,0) to (6,6), using Pythagorean theorem, is 6sqrt(2).
But, I don't think you know anything about the sides of these triangles, other than one side is 6.

4. Re: Determine the co-ordinates

Allow me to clarify (and I sure wish soroban were here with his sweet ASCII art!...)

Somewhere above (0, 6) on the y-axis, we have (0, b) - an endpoint of the segment.
Similarly, (a, 0) is to the right of (6, 0)
The triangle with vertices (0, 6), (6, 6) and (0, b) - with legs of length 6 and (b - 6)
is similar to the triangle with vertices (6, 6), (a, 0) and (6, 0) - with legs of length 6 and (a - 6)

You should not be calculating the diagonal of the square!

So using proportions,

$\displaystyle \frac{b - 6}{6} = \frac{6}{a - 6}$

Hmm... I'm not doing scratchwork, so it looks like the Pythagorean theorem route isn't the best option. There might be a way to go directly to a relationship between a and b, in a linear equation.
Then we combine this with our proportions.
IDK... off to church!

5. Re: Determine the co-ordinates

Originally Posted by EastCan
Thanks for your reply. What you've said, is what I have done. The rectangle (square) is 6 by 6, and the diagonal from (0,0) to (6,6), using Pythagorean theorem, is 6sqrt(2).
You know that $\displaystyle a^2+b^2=21$.
You also know about slope: $\displaystyle \frac{6-0}{6-a}=\frac{6-b}{6-0}$.

6. Re: Determine the co-ordinates

Have you forgotten everything?

x, and y axis. Points (0,r) and (s,0)

Equation of this line is $\displaystyle \frac{x}{s}+\frac{y}{r}\;=\;1$

Length is 21. $\displaystyle r^{2} + s^{2} = 21^{2}$

(6,6) is on there. $\displaystyle \frac{6}{s}+\frac{6}{r}\;=\;1$

It's a good, symmetric problem. Two solutions should present themselves.

7. Re: Determine the co-ordinates

Have I forgotten everything? Ouch! That hurts.

8. Re: Determine the co-ordinates

I understand what Plato has written. TKHunny, can you kindly help me out with your equation of the line. I must have forgotten alot apparently (thanks for driving that point home), because I don't see immediately how you came up with that equation.

I'm used to the y=mx+b form of a line, so I'm probably getting hung up on that.

Looking through some math references, I understand that form of equation that you use.

Thank you for your help thus far, I will work on this a bit and see what I come up with.

9. Re: Determine the co-ordinates

Originally Posted by EastCan
TKHunny, can you kindly help me out with your equation of the line.
If $\displaystyle (a,0)~\&~(0,b)$ are the intercepts and $\displaystyle ab\ne 0$ then the equation of that line is $\displaystyle \frac{x}{a}+\frac{y}{b}=1$.

10. Re: Determine the co-ordinates

This is a pretty old problem, I remember being given it ages ago. Then it appeared in the form of trying to get into a girls bedroom window. Her father had built a 6ft wall 6ft away from the building and the problem was to find out whether a 21ft ladder was long enough to reach the window, or something like that.
Anyway, from a maths point of view, suppose that the intercepts on the x and y axes are a and b, then
$\displaystyle a^2+b^2=21^2,$ which can be rewritten in the form $\displaystyle (a+b)^2-2ab=21^2.$
The equation of the line (the ladder) will be $\displaystyle x/a+y/b=1,$ and since it passes through the point (6,6), $\displaystyle 6/a+6/b=1$ from which $\displaystyle (a+b)/ab=1/6.$
Use this to substitute for $\displaystyle ab$ in the earlier equation and you have a quadratic in $\displaystyle a+b$ which you solve.
So, find b in terms of a and substitute into the original Pythagoras equation.

11. Re: Determine the co-ordinates

Hi BobP, thank you for your help. It is very helpful. I was working on this, and the algebra gets very messy when trying to solve the quadratics. Do you think this is to be expected, or should the answer to this problem have a neat little answer?
I will have another go at it, in any case.

12. Re: Determine the co-ordinates

I didn't think that it was particularly messy, the first quadratic is$\displaystyle (a+b)^2-12(a+b)-441=0$ from which $\displaystyle a+b\approx 27.84033.$

I got 19.09 and 8.75 (2dp) for the values of a and b (or b and a).

13. Re: Determine the co-ordinates

I was assuming I was not permitted to reduce the answers to decimal form, for whatever reason. Otherwise, it is easy to solve at this point. Thank you very much for your help.

14. Re: Determine the co-ordinates

The answers in exact form are

$\displaystyle 3\;\frac{\sqrt{53}+2-\sqrt{41-4\sqrt{53}}}{2}\quad \text{ and } \quad 3\;\frac{\sqrt{53}+2+\sqrt{41-4\sqrt{53}}}{2}$.