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## Co-ordinate Geometry - Innovative Question on Rotation

If you have run through the Co-ordinate Geometry module, then you must check out this question. It may be beyond GMAT level but it tests the GMAT relevant concepts  in a most innovative way.

Question: Triangles ABC and ADE have areas 2007 and 7002, respectively, with B = (0, 0), C = (223, 0), D = (680, 380), and E = (689, 389). What is the sum of all possible x-coordinates of
A?

(A) 282
(B) 300
(C) 600
(D) 900
(E) 1200

Solution:

The worse the numbers given, the more the probability that they will not get used at all.
Draw a diagram. BC lies on X axis. To make a triangle with a given area (2007) i.e. a fixed altitude (whatever that may be, it is 18 in this case), A will lie on any point on the two red lines. The two red lines are 18 units above and below the x axis. Think about this before moving ahead.

Now draw DE. To make a triangle with a given area (7002) i.e. a fixed altitude (whatever that may be), A will lie on any point on the two given dotted purple lines.

So what we need is the x co-ordinates of the 4 possible A values (A', A'', A''' and A'''')

Now notice the co-ordinates of D and E are D = (680, 380), and E = (689, 389). So the diff between x co-ordinates is 9 and between y co-ordinates is also 9. This means DE is a line with slope 1.
Hence, it will cut the x axis at 300 (when we move y co-ordinate 380 units down, we will move the x co-ordinate 380 units to the left so the x co-ordinate will become 680 - 380 = 300). This is the point M.

Now note that slope of purple dotted lines is also 1 since they are parallel to DE.
By symmetry, the 4 possible x co-ordinates of A
A' = 300 + a + b
A'' = 300 + a - b
A''' = 300 - (a + b)
A'''' = 300 - (a - b)

When we add them, all a and b will get cancelled to give a total of 1200.