sachindia wrote:What are the coordinates for the point on Line AB that
is three times as far from A(-5,6) as from B(-2,0) and that is in between points
A and B?

Since the point divides A,B into 3:1, the point is:-

(3B+1A)/(3+1) = 1/4 * ( 3(-2,0) + 1(-5,6)) = 1/4 ( (-6,0)+(-5,6)) = 1/4 ( -11, 6) = (-11/4, 6/4) = (-2.75, 1.5)

Thanks eagleeye,

but I dint understand

Since the point divides A,B into 3:1, the point is:-

(3B+1A)/(3+1)

how (3B+1A)/(3+1) ?????

sachindia wrote:Thanks eagleeye,

but I dint understand

Since the point divides A,B into 3:1, the point is:-

(3B+1A)/(3+1)

how (3B+1A)/(3+1) ?????

if a point C=(x,y) divides A:B in m:n ratio, then the point is:

(x,y) = (mB+nA)/(m+n). Just remember this formula. Here's how I derived it:
Since C divides AB in m:n ratio
AC/CB = m/n
(C-A)/(B-C) = m/n
=> nC-nA = mB-mC
=> (m+n)C = mB+nA
=> C = (mB+nA)/(m+n)

In the case above m=3, n=1.

Did this help ?

Question for you Eagleeye, but why does the equation equal 1/4--> ((3B+1A)/(3+1) = 1/4). Is it because we want to find the distance that is 3 times as far from A as it is from B so the total distance between point A and B would be 4 or am I just missing something?


Thanks

tisrar02 wrote:Question for you Eagleeye, but why does the equation equal 1/4--> ((3B+1A)/(3+1) = 1/4). Is it because we want to find the distance that is 3 times as far from A as it is from B so the total distance between point A and B would be 4 or am I just missing something?


Thanks

tisrar02:

I am not sure what you are asking here, but I've derived the formula in the previous post. The equation does NOT equal 1/4.

We need to find the co-ordinates of point C that divides A and B in the ratio 3:1.

So I wrote C = (3B+1A)/(3+1) = 1/4*(3B+1A) = 3B/4 + A/4.

I feel like I'm not getting what you're asking. Or did this answer your query?

Sorry for my confusing question,

I was wondering as to where the 1/4 came from

For this sort of problem, you can usually determine the correct answer quickly by process of elimination.

Check my posts here:

https://www.beatthegmat.com/coordinate-g ... 58176.html

https://www.beatthegmat.com/grockit-tough-problem-t74452.

tisrar02 wrote:Sorry for my confusing question,

I was wondering as to where the 1/4 came from

Ok, I will derive this once again:

Let the point we need to find be C.

Now, C divides AB in the ratio 3:1, Hence AC:CB = 3:1
AC/CB = 3/1
AC = 3CB
C-A = 3(B-C)
C-A = 3B-3C
3C+C = 3B+A
4C = 3B+A
C = 1/4*(3B+A).

So that's where the 1/4 comes from.

(By the way, my derivation makes use of vector algebra which is not tested by the GMAT, I just derived it here using vectors for simplicity). GMAT would expect you to use similar triangles to do the same.

If you ever face a similar question on the GMAT, use the formula.

If AC:CB = m:n
C = (mB+nA)/(m+n)

This also tells us the special case that mid-point of a line segment AB = (A+B)/2.

Are we on the same page now?

Eagleeye,

Yes, I understand now. Thank you for the explanation. I always try to figure out problems using the harder way and the easier ways so both you and Mitch really helped me out!

Thanks

to get coordinates.

Divide the difference in y-coordinate i.e (6-0) into 4 parts (3:1) i.e 3/2

Divide the difference in x-coordinate i.e (-5+2) into 4 parts i.e -3/4

To get the actual coordinates, [-2-3/4 , 0+3/2] = [-2.75,1.5)