Coordinate Geometry!

[sachindia]

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?

[eagleeye]

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)

[sachindia]

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) ?????

[eagleeye]

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 ?

[tisrar02]

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

[eagleeye]

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?

[tisrar02]

Sorry for my confusing question,

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

[GMATGuruNY]

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.

[eagleeye]

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?

[tisrar02]

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

[anjalimanas]

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)