I got the math correct by using few calculations ( or right triangles and then mid point). But it took much time.
Is their any GRE-way to save time in such case?
Thank you.

I think no. Above is the fastest solution that I know.

Maybe there is another approach. Maybe @GreenLightTestPrep could come in handy...............

But how point (8, 3) could lie in a same line while we may calculate this line's equation which is y = - 3/4 * x + 9?

Only points B and E lie there!

what of point (1,10)

First sketch the given information:



Since the slope (rise/run) of the line is -3/4, for every 3 units we move UP, we move 4 units to the LEFT (alternatively, we can say for every 3 units we move DOWN, we move 4 units to the RIGHT)

Notice that we end up with a RIGHT triangle with legs of length 3 and 4, which means the hypotenuse must be length 5.
So, the point (0, 9) is on the line AND it is 5 units from the point (4, 6)


Likewise, if we start from (4, 6) and move 3 units DOWN, and 4 units to the RIGHT, we get the following:

Once again, we end up with a RIGHT triangle with legs of length 3 and 4, which means the hypotenuse must be length 5.
So, the point (8, 3) is on the line AND it is 5 units from the point (4, 6)


Answer: C, E

Cheers,
Brent

The point (1, 10) is, indeed, 5 units from (4, 6). However, (1, 10) does not lie ON the given line.

Cheers,
Brent

A = (4, 6)
B = (x, y)

Since, they both lie on the same line, they must have same slope as \(\frac{-3}{4}\)

A. slope = \(\frac{(6 - 1)}{(4 + 1)} = 1\)

B. slope = \(\frac{(6 - 12)}{(4 + 4)} = \frac{-3}{4}\)

C. slope = \(\frac{(6 - 3)}{(4 - 8)} = \frac{-3}{4}\)

D. slope = \(\frac{(6 - 10)}{(4 - 1)} = \frac{-4}{3}\)

E. slope = \(\frac{(6 - 9)}{(4 - 0)} = \frac{-3}{4}\)

Now, lets check the distance!

B. \((6 - 12)^2 + (4 + 4)^2 ≠ 5^2\)

C. \((6 - 3)^2 + (4 - 8)^2 = 5^2\)

E. \((6 - 9)^2 + (4 - 0)^2 = 5^2\)

Hence. option C and E

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