Question 178260
If we draw a picture, we get

<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/Algebra%20dot%20com/step1.png">

Now let the horizontal distance from the poles be the value "a" (this variable is needed, but as you will soon see, it does not change the final answer). Also, label the top and bottom of both poles with coordinates (draw some coordinate axis  if necessary)



<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/Algebra%20dot%20com/step2.png">





So to find the height of the point where the ropes cross, we first need to find the equations of the lines (the red and blue lines)



Equation of the first line from (0,12) to (a,0) (red line)



Slope: {{{m=(y[2]-y[1])/(x[2]-x[1])=(12-0)/(0-a)=12/(-a)=-12/a}}}



So the slope of the red line is {{{m=-12/a}}}



{{{y=mx+b}}} Start with the slope-intercept general equation



{{{12=(-12/a)(0)+b}}} Plug in {{{m=-12/a}}}, {{{x=0}}} and {{{y=12}}} (since the red line goes through the point (0,12))



{{{12=0+b}}} Multiply



{{{12=b}}} Simplify



So the first equation (the equation of the red line) is {{{y=(-12/a)x+12}}}




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Equation of the second line from (0,0) to (a,6) (blue line)



Slope: {{{m=(y[2]-y[1])/(x[2]-x[1])=(0-6)/(0-a)=(-6)/(-a)=6/a}}}



So the slope of the blue line is {{{m=6/a}}}



{{{y=mx+b}}} Start with the slope-intercept general equation



{{{0=(6/a)(0)+b}}} Plug in {{{m=6/a}}}, {{{x=0}}} and {{{y=6}}} (since the blue line goes through the point (0,0))



{{{0=0+b}}} Multiply



{{{0=b}}} Simplify



So the second equation (the equation of the blue line) is {{{y=(6/a)x}}} 




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Now equate the two equations {{{y=(-12/a)x+12}}} and  {{{y=(6/a)x}}} 


{{{(6/a)x=(-12/a)x+12}}} 



{{{6x=-12x+12a}}} Multiply EVERY term by the LCD "a" to clear the fractions.



{{{6x+12x=12a}}} Add 12x to both sides.



{{{18x=12a}}} Combine like terms.



{{{x=(12a)/18}}} Divide both sides by 18 to isolate "a".



{{{x=(2/3)a}}} Reduce. Note: this tells us that no matter what the value of "a" is, the crossing point will ALWAYS be {{{2/3}}} of the horizontal length




{{{y=(6/a)x}}} Go back to the second equation



{{{y=(6/a)((2/3)a)}}} Plug in {{{x=(2/3)a}}}



{{{y=(12a)/(3a)}}} Multiply



{{{y=4}}} Reduce



So the point of intersection of the red and blue lines is *[Tex \LARGE \left(\frac{2}{3}a,4\right)] for any value of "a"




Since the y-coordinate is 4, this means that the height of the crossing point is 4 ft.




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Answer:


So the ropes cross 4 ft above the ground regardless of the horizontal distance between the two poles.