Pre-calc application problem

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Sam

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The problem:
You are playing a game of billiards on a standard 92" by 46" pool table. In this game you score points by bouncing the ball off as many sides of the table as possible prior to sinking the ball in the called pocket. The ball is currently 10 inches down and 12 inches to the right of the top left pocket. You call that you will sink it in the bottom right corner pocket. Write functions to determine the path of the ball if:

a) Sink it without bouncing off any sides.
b) Sink it bouncing off exactly one side.
c) Sink it bouncing off exactly two sides.
d) Sink it bouncing off a large number of sides.

I have made a graph of the pool table where the origin is at the middle so the domain of the graph is [-23,23] and the range is [-46,46]. I have found the equation for part A but I am not sure how to get the absolute value equation to hit the 2 points( the ball, (-11,46) and the bottom right corner pocket (23,-46). What I have now is y= -|x|+46, I'm am not sure what to scale it by.

Thanks, Sam
 
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The "first method" at the site Denis links to is specifically for situations in which the distances are equal. That is not the case here. Setting up the coordinate system as you say, the ball is at (-11, 36) and the pocket is at (23, 0). The equation of the straight line from the first to the second would be easy to get. For (b) you want to bank the ball off the opposite side which has equation x= -23. Let Y be the y coordinate of the point where the ball strikes that side. The line from (-11, 36) to (-23, Y) is \(\displaystyle y= \frac{36-Y}{12}(x+ 11)+ 36\). The line from (-23, Y) to (23, 0) is \(\displaystyle y= \frac{Y}{-46}(x- 23)\). The angles those lines make with the side, x= -23, are the complements of the anges they would make with a line of constant y. The angles they make with x= -23 are the same, then the angles they make with y= constant will be the same which means that the slopes of the lines will be the same: \(\displaystyle \frac{36- Y}{12}= \frac{Y}{-46}\). Solve that for Y.
 
HallsofIvy said:
The "first method" at the site Denis links to is specifically for situations in which the distances are equal. That is not the case here. Setting up the coordinate system as you say, the ball is at (-11, 36) and the pocket is at (23, 0). The equation of the straight line from the first to the second would be easy to get. For (b) you want to bank the ball off the opposite side which has equation x= -23. Let Y be the y coordinate of the point where the ball strikes that side. The line from (-11, 36) to (-23, Y) is \(\displaystyle y= \frac{36-Y}{12}(x+ 11)+ 36\). The line from (-23, Y) to (23, 0) is \(\displaystyle y= \frac{Y}{-46}(x- 23)\). The angles those lines make with the side, x= -23, are the complements of the anges they would make with a line of constant y. The angles they make with x= -23 are the same, then the angles they make with y= constant will be the same which means that the slopes of the lines will be the same: \(\displaystyle \frac{36- Y}{12}= \frac{Y}{-46}\). Solve that for Y.
Click to expand...

Thanks for the help.
I got y=36.8 but when I plug it back in I don't get the same answers. Also the pocket is at (23,-46).
 
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