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

Can you please explain it?

OE


Using the definition of slope as \(m=\frac{rise}{run}=\frac{3}{4}\) you can plot point A and move vertically 3 and horizontally − 4 to point (4 − 4, 6 + 3) = (0, 9), which will also lie on the line with slope \(- \frac{3}{4}\)

This creates a 3: 4: 5 triangle, so the distance along the line from point A to the new point (0, 9) is 5 units, so (E) could be point B. You can also move vertically − 3 and horizontally 4 to point (4 + 4, 6 − 3) = (8, 3), which lies on the same line. Since the triangle formed is a 3: 4: 5 triangle again, this distance from point A to the new point (8, 3) is also 5. Choice (C) could also be point B. So, the answers are (C) and (E).

Please also refer to the explanations above by the GRE tutors

1) https://gre.myprepclub.com/forum/point- ... tml#p35297
2) https://gre.myprepclub.com/forum/point- ... tml#p65973