Coordinate Geometry!

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sachindia: Master | Next Rank: 500 Posts; Posts: 258; Joined: Fri Jun 22, 2012 4:39 amFollowed by: 3 members; Location: Bengaluru, India; Thanked: 6 times; GMAT Score:640

by sachindia » Sat Aug 04, 2012 4:57 am

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?

Regards,
Sach

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Source: Beat The GMAT — Problem Solving | Sat Aug 04, 2012 4:57 am

eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members; Thanked: 339 times; GMAT Score:770

by eagleeye » Sat Aug 04, 2012 5:10 am

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)

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sachindia: Master | Next Rank: 500 Posts; Posts: 258; Joined: Fri Jun 22, 2012 4:39 amFollowed by: 3 members; Location: Bengaluru, India; Thanked: 6 times; GMAT Score:640

by sachindia » Sat Aug 04, 2012 5:16 am

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) ?????

Regards,
Sach

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eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members; Thanked: 339 times; GMAT Score:770

by eagleeye » Sat Aug 04, 2012 5:35 am

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 ?

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tisrar02: Master | Next Rank: 500 Posts; Posts: 130; Joined: Fri Apr 20, 2012 8:13 amFollowed by: 4 members; Location: Toronto, Ontario; Thanked: 16 times; GMAT Score:650

by tisrar02 » Sat Aug 04, 2012 5:56 am

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

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eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members; Thanked: 339 times; GMAT Score:770

by eagleeye » Sat Aug 04, 2012 6:10 am

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?

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tisrar02: Master | Next Rank: 500 Posts; Posts: 130; Joined: Fri Apr 20, 2012 8:13 amFollowed by: 4 members; Location: Toronto, Ontario; Thanked: 16 times; GMAT Score:650

by tisrar02 » Sat Aug 04, 2012 6:17 am

Sorry for my confusing question,

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

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Aug 04, 2012 6:24 am

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

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eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members; Thanked: 339 times; GMAT Score:770

by eagleeye » Sat Aug 04, 2012 6:34 am

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?

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tisrar02: Master | Next Rank: 500 Posts; Posts: 130; Joined: Fri Apr 20, 2012 8:13 amFollowed by: 4 members; Location: Toronto, Ontario; Thanked: 16 times; GMAT Score:650

by tisrar02 » Sat Aug 04, 2012 6:37 am

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

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anjalimanas: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Fri Apr 27, 2012 1:49 am

by anjalimanas » Wed Nov 21, 2012 3:16 am

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)

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