On the xy coordinate plane, point C is (5,-2) and point D is

Well if you have issues in finding the line equations. You can find it using the straightforward formula. If point p devices the line segment in the ratio m:n in this case m:n is 2:1. The 2 is the part closer to c.
Hence the formula is
X(new) = (mx1 + nx2 )/( m+n)
Similarly for y
Y(new) = (my1 + ny2 )/( m+n)
Here I take ( x1,y1) as the coordinates of D and (x2,y2) as the coordinates of C.
So X(new)= (-2+5)/(2+1)= 1
Similarly Y(new)= (-2+5)/(3)=1
Hence ans is B.

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Hope it helps.

Hi dear math experts, it this question type in scope of gmat ? I just don't want to memorize one more formula, as there are plenty of them in geometry.

Hi BrainLab,

Yes, this question is certainly within the scope of the GMAT. The good news is that you don't need to memorize any formulas to solve it. It can be with a little logic and a few simple calculations. It's also ALWAYS a good idea to sketch the line to be clear on where the points are and what you're looking for.

We want the point on the line that is twice as far from C as it is from D. We can examine the x and y coordinates separately since they are independent.

*It should be noted that there are two solutions to this problem, one point between C and D, and another point with D in the middle of C and the point. We can quickly look at the answer choices and see that all the points are between C and D, therefore we can search for that point using the following method:

Taking the x-coordinate first, the distance between C and D is |(x-coordinate of C) - (x-coordinate of D)|= |5 - (-1)| = 6
The x-coordinate that is twice as far from C as it is from D (and in between C and D) will be 4 units from C and 2 units from D. So the x coordinate of the point we're looking for is 5-4 = 1, or (-1)+2 = 1.

Now looking at the y-coordinate, we do the same thing. The distance between C and D is |(-2) - 2.5| = 4.5
The y-coordinate that is twice as far from C as it is from D (and in between C and D) will be 3 units from C and 1.5 units from D. So the y coordinate of the point we're looking for is (-2)+3 = 1, or 2.5-1.5 = 1.
So now we have our coordinates of the point on the line that is twice as far from C as it is from D: (1,1)

Answer: B

*Note: If we were asked to find the other point on the line that is twice as far from C as it is from D, then instead of dividing the distance between the points by 3, we would simply add/subtract the distance to/from the coordinates of D, in the direction away from C:
x-coordinate: x distance between C and D is 6, so x-coordinate will be (-1)-6 = -7
y-coordinate: y distance between C and D is 4.5, so y-coordinate will be 2.5+4.5 = 7
So point (7,7) is also on the line and twice as far from C as it is from D, but is obviously not one of our answer choices.

Cheers

On the xy-coordinate plane, point C is (5;?2) and point D is
(?1; 2:5). The point on line segment CD that is twice as far
from C as from D is
(A) (1;?1)
(B) (1; 1)
(C) (2; 0:25)
(D) (3; 0:5)
(E) (3; 1)

Merging topics. Please refer to the discussion above.

point C is (5,-2) and point D is (-1,2.5).

The point on line segment CD that is twice as far from C as from D is:

We have to figure out the point that divides the line segment in 2:1 ratio therefore

(2*-1+5)/3,(5-2/3) through(new point formation through the ratio)

Hence IMO (-1,-1) or rather B