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by Bill@VeritasPrep » Sat Mar 31, 2012 5:17 pm
We can use the two known points to find the slope of the line:

=(y2 - y1)/(x2 - x1)
=(3 - 2)/(5 - 3)
=1/2

In other words, we move up 1 unit for every 2 units we move right.

Since (x,4) is one unit up from (5, 3), x must be 2 units to the right of 5, which gives us a value of 7.
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by gmatmath » Fri Apr 06, 2012 8:34 am
if (5,3), (x,4) and (3,2) lie on the same line, they are collinear.
This means that the slope of any 2 points taken should be the same.

1)(5,3) and (x,4)
slope = (4-3)/(x-5)

2) (5,3) and (3,2)
slope = (2-3)/(3-5)

Now, equating both the slopes


1/(x-5) = -1/-2
==> 1/(x-5) = 1/2
since the numerators are equal, we can equate the denominators.

So, x-5 = 2
==> x = 7
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by neelgandham » Fri Apr 06, 2012 9:46 am
As mentioned by Bill and gmatmath, Slope formula works best.

But here is something I do,
if I have no clue,
how to do - ah it rhymes!
(Not only to solve this problem, but to solve any geometry problem). We all know that the erase pad provided in the GMAT exam centre contains grids. So, I use the grid lines to draw a Cartesian coordinate system and plot the points in the system. For example, in the question above, I plotted the points (5,3) and (3,2). I, then extended the line segment to meet the line y = 4 at (7,4). and the rest...

p.s: Use this as the last resort[/i}
The attachment is self-explanatory.
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