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GMAT85
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Post subject: CAT 3:The Incredible line (altenate Soln.)  Posted: Sat Jul 12, 2008 10:37 am |
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In the xy-coordinate system, what is the slope of the line that goes through the origin and is equidistant from the two points P = (1, 11) and Q = (7, 7)?
MGMAT answer:
Let's call R the midpoint of the line segment between P and Q. R's coordinates will just be the respective average of P's and Q's coordinates. Therefore R's x-coordinate equals 4 , the average of 1 and 7. Its y-coordinate equals 9, the average of 11 and 7. So R=(4, 9).
Finally, the slope from the (0, 0) to (4, 9) equals 9/4, which equals 2.25 in decimal form.
My Answer:
The way I would do this question is form a line joining the points P and Q. So the negative reciprocal of the slope PQ should be the answer to this question. Slope of line PQ: (11-7)/(1-7) = 4/-6. The negative reciprocal of this is 6/4 or 3/2.
I don't understand why my approach doesn't work on this question???
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Guest
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Post subject: Posted: Sun Jul 13, 2008 12:01 am |
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simple answer:
its not necessary for the lines to be perpendicular to each other.
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Guest
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Post subject: Posted: Sun Jul 13, 2008 12:04 am |
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as a follow up:
say the points are (5,2), (11,2) (line parallel to x-axis. The mid-point is 8,2
But according to you, the new line should be parallel to y-axis. Is it?
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RonPurewal
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Post subject: Posted: Sat Jul 19, 2008 12:28 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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well played, "guest".
incidentally, although the line through R is equidistant from P and Q, the distance between the line and P (or Q) is NOT the same as the distance from R to P (or R to Q).
if anyone is interested, i can upload a hand drawn diagram to explain this little nicety.
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TP
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Post subject: Posted: Sun Aug 10, 2008 12:02 am |
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RPurewal wrote: well played, "guest".
incidentally, although the line through R is equidistant from P and Q, the distance between the line and P (or Q) is NOT the same as the distance from R to P (or R to Q).
if anyone is interested, i can upload a hand drawn diagram to explain this little nicety.
Ron,
I had a question and a comment regarding this question:
Q. I "saw" another line, that goes through the origin, parallel to the line P-Q. This might sound silly, but
isn't this new line equidistant from both P & Q ?. We can drop perpendiculars from P and Q to this new
line and as it is parallel to line P-Q, the distances would be the same?
Comment: Although I did seem to understand Guest's reply, I am unable to visualize it. May I take
you up on the offer for a quick hand drawn diagram?
thanks.
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TP
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Post subject: Posted: Sun Aug 10, 2008 12:07 am |
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TP wrote: Comment: Although I did seem to understand Guest's reply, I am unable to visualize it. May I take
you up on the offer for a quick hand drawn diagram?
thanks.
Ignore the comment :) I was able to visualize it. Still have the question tho.
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TP
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Post subject: Posted: Wed Aug 13, 2008 3:11 am |
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RPurewal wrote: well played, "guest".
incidentally, although the line through R is equidistant from P and Q, the distance between the line and P (or Q) is NOT the same as the distance from R to P (or R to Q).
if anyone is interested, i can upload a hand drawn diagram to explain this little nicety.
Ron,
I uploaded a rough sketch for my question.
http://www.postimage.org/image.php?v=Pq1YTpI9
Wouldn't you consider line B to be equidistant
from the 2 points (1, 11) and (7, 7) on line A ?
I am sure there is something really silly about what I am saying.
Thanks
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Guest
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Post subject: Posted: Mon Sep 01, 2008 3:12 pm |
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Hello Everybody,
This is exactly I tried to solve : http://www.postimage.org/image.php?v=Pq1YTpI9
and I couldn't get to the answer. What am I doing wrong ? Doesn't this question has 2 solutions ? Please point out the mistake. Thanks.
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Guest
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Post subject: Posted: Tue Sep 02, 2008 6:06 pm |
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Could someone please respond to the previous question ?
Hello Everybody,
This is how exactly I tried to solve : http://www.postimage.org/image.php?v=Pq1YTpI9
and I couldn't get to the answer. What am I doing wrong ? Doesn't this question has 2 solutions ? Please point out the mistake. Thanks.
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Guest
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Post subject: Posted: Mon Sep 15, 2008 12:06 am |
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Could anybody please respond ? Even I was tricked by the same question and I drew the same diagram as people drew here in the post.
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RonPurewal
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Post subject: Posted: Thu Oct 09, 2008 7:24 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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yes, the alternate solution posted here is correct; the line parallel to the original line is, indeed, equidistant from those two points.
well played.
we will fix the problem. one really easy way to fix the problem is to state that the line bisects segment PQ rather than merely stating that it's equidistant from the two points.
thanks.
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TP
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Post subject: Posted: Sat Oct 11, 2008 2:35 am |
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RPurewal wrote: yes, the alternate solution posted here is correct; the line parallel to the original line is, indeed, equidistant from those two points.
well played.
we will fix the problem. one really easy way to fix the problem is to state that the line bisects segment PQ rather than merely stating that it's equidistant from the two points.
thanks.
Thank you Sir Ron! :). You are a good man!
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idiot
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Post subject: there can be two such lines Posted: Fri Oct 17, 2008 11:17 am |
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actually it's possible to draw two such lines-
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esledge
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Post subject: Posted: Mon Oct 27, 2008 6:36 pm |
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| ManhattanGMAT Staff |
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Posts: 901
Location: St. Louis, MO
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Idiot's picture is a correct and complete reason that there are two solutions to the problem as written.
I agree with Ron that a good fix would be to specify that the line bisects segment PQ. In fact, just saying that the line intersects PQ would be sufficient to rule out the parallel-to-PQ case, and would be less likely to unfairly bias people into thinking the line must be perpendicular to PQ, but either wording would be mathematically valid.
_________________
Emily Sledge
Instructor
ManhattanGMAT
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