This is kind of a tricky question, one I doubt you'd see (though you never know).. There are several different ways to solve it (maybe there are more that I don't see). If someone knows a better method, post it! I'll try to break down the method described by rockclock best I can and also offer an alternative. Note: Try drawing the steps described below out on a coordinate plane as it will make them much easier to follow.
Method 1: Just by looking at the picture, we know that for every 2 units we go over in the x direction, we go down (or up) 3 units in the y direction. So the slope (rise divided by run) is +/- 3/2. It is negative when going down and positive when going up.. Since it is a reflection, we know that the magnitude of the slope will be the same. So starting at (0,3) i.e. the Y axis, going down 3 and over 2 puts us at (2,0), which is also shown in the image. Now, the beam is going to be reflected upwards until it reaches the ceiling of the square (which is 10 units up). So starting at (2,0) and going up 3 units for every 2 units we go to the right, we go to (4,3) [from the slope (2+2, 0+3)].. Following this same logic, the light continues upwards from (4,3)-->(6,6)-->(8,9). This is where it gets a little tricky. If we were to go up another 3 units and over 2, that would put us at (10, 12) which is outside the confines of the square (remember it is a 10 x 10 square). To get up to a Y coordinate of 10 from 9, we are only going one unit upwards. Even though we have expressed the slope as 3/2 , we could just as easily express it as 6/4, 8/6.. or even (2/3)/1 as in "two-thirds divided by 1." These ratios are all mathematically the same. In other words, for every 1 unit we go over to the right, we go up 2/3 units. I chose to look at it this way because we are going 1 unit up to get from 9 to 10. This 1 unit vertical increase is 1/3 of 3. For the slope ratio to be accurate, we must also take 1/3 of the horizontal component of the slope (recall that this was 2). 1/3 of 2 is 2/3. Thus, going over 2/3 units from 8 puts us at an X coordinate of 26/3. At this point, we are now at the ceiling of the square (26/3, 10). The light beam will be reflected downwards and to the right until we reach the right wall (which is also at 10). To get from an X coordinate of 26/3 to an X coordinate of 10, we are going 4/3 units to the right (10 minus 26/3 = 4/3). Using the same logic as before, 4/3 is 2/3 of the horizontal movement to the right (since the horizontal component of slope, 4/3 divided by 2 = 2/3). And thus, the vertical movement downwards will be 2/3 OF 3 (the vertical component of slope) which gives us 2 (since 2/3 of 3 is simply (2/3) x 3 = 2). So to take a step back, what I am doing is expressing the slope so that we know exactly where each coordinate will be when it reaches the wall. So for this reflection, we are going down 2 for every 4/3 we move to the right. Remember that we are still at the ceiling at (26/3, 10). So going 4/3 to the right and down 2 puts us at (26/3 + 4/3, 10-2) = (10,8), which is on the right wall. The final reflection will end up sending us to the left and downwards. So from (10,8), we will move down 3 for every 2 we move to the left. So will go from (10,8)-->(8,5)-->(6,2) and finally to (14/3, 0), which is on the X axis and is our answer. We get the 14/3 using the same steps as we did before.
I know that there are a lot of fractions and details in this explanation. Even though it is extensive, the problem really doesn't take too long once you understand the concepts and can execute them rather quickly (no more than one minute).
Method 2: You can write equations for each line using y = mx + b (slope-intercept form). Remember that m is slope and b is the Y intercept. Again, tracing out the lines makes it easier to follow. Starting at the bottom, (2,0), we can write an equation for the line going up and to the right until it reaches the ceiling. Using slope intercept form, the equation for the line is y = (3/2)x - 3. To get that, all you need is the slope of the line and one point that lies on the line. We know that (2,0) lies on that line and the slope is 3/2. Let's plug them in the y = mx + b equation to solve for b and then write our equation. In this case, we will use the X coordinate of 2 and the Y coordinate of 0. So (0) = (3/2)(2) + b. Solving for b gives us b = -3. Thus, m = 3/2 and b = -3 and our equation is y = 3/2x - 3. Now, in order to get the x coordinate at the ceiling, let's plug y = 10 (since we know the y coordinate at the ceiling is 10) into our linear equation and solve for X.. 10 = (3/2)x -3. This gives us X = 26/3, which is what we got in method 1 also. Now do the same for the next reflection. The next line will move down and to the right until it hits the right wall. Using our newly found ceiling point (26/3, 10) and a slope of -3/2 (slope is now negative since the line slopes downwards) to solve for b, our new equation is y = -3/2x + 23. Plugging the right wall's x coordinate (x = 10) into the equation, we get a y coordinate of 8. We can now use this (10,8) point to derive our final equation. Using the (10,8) point and a slope of 3/2 (it's now positive again because it slopes upwards mathematically) we get an equation of y = (3/2)x - 7. Since they want the X coordinate where the beam reaches the bottom (where the y coordinate equals zero), we plug in y = 0 into our final equation and solve for X. 0 = (3/2)X -7. This gives us X = 14/3
I hope that clears things up. If it didn't, please ask away
