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Stuart Kovinsky wrote:

ricky wrote:This is exactly what i was thinking Ian.The ques seems wrong to me as well.Surprisingly It is Manhattan GMAT Test !

Who is this Ian of whom you speak!

I'm guessing it's me.

Stuart Kovinsky wrote:
The only way that a "line" can be equidistant from two points is if it's a perpendicular bisector of the line joining those two points.

A line which is equidistant from two points does not need to be a perpendicular bisector. It only needs to be a bisector (edit- unless the line is parallel to the line joining the two points). Consider the following two points:

(-3, 1) and (3, -1)

The x-axis is clearly equidistant from both points (we measure distances at right angles, and the x-axis is one below the first point, and one above the second). But while the x-axis bisects the line joining (-3, 1) and (3, -1), it is not perpendicular to that line.

It's not difficult to see why we only need the midpoint in the given question, but it's more difficult to demonstrate without a diagram. Draw two points A and B in quadrant I of the coordinate plane. Connect A and B, and label the midpoint M. Draw the line containing the origin and M. To measure the distance from A to this line, we draw a right-angled triangle connecting A to the line, with hypotenuse AM. This triangle is exactly congruent to the right-angled triangle you get when you measure the distance from B to the line (the angles are equal, and AM=BM). We don't need a perpendicular bisector; we only need the midpoint.

As has been established above, the midpoint of (1, 11) and (7, 7) is (4, 9), and the slope of a line through both (0,0) and (4,9) is 9/4. There is still a problem with the question, however. A line which is parallel to the line connecting (1,11) and (7,7) will also be equidistant from both points. So there are two possible answers, 9/4 and -2/3, and the question is wrong to speak of the line which is equidistant from the points; it should ask for the slope of a line which is equidistant.