BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

slope of the line

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 575
Joined: Tue Nov 04, 2008 2:58 am
Location: India
Thanked: 18 times
Followed by:4 members
GMAT Score:710

slope of the line

by rahulg83 » Sun Mar 01, 2009 3:50 am
This one is a 700-800 problem from one of the manhattan online tests:

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)?

A) 2
B) 2.25
C) 2.50
D) 2.75
E) 3

My doubt is if the line is equidistant from the mentioned points, does that means mid-point of the line segment joining the two points P and Q lies on the line passing through origin. I am bit confused as we calculate distance of a point from by dropping a perpendicular onto the line.

OA later
Top
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 151
Joined: Wed Feb 11, 2009 2:45 am
Thanked: 3 times

by deepoe » Sun Mar 01, 2009 4:01 am
I found this maybe it will help:

https://mathforum.org/library/drmath/view/54426.html
Top
Master | Next Rank: 500 Posts
Posts: 392
Joined: Thu Jan 15, 2009 12:52 pm
Location: New Jersey
Thanked: 76 times

by truplayer256 » Sun Mar 01, 2009 6:11 am
I'm not too sure about my answer but I think that the answer is B.

The only way the line is going to be equidistant from Point P and Point Q is if it passes through the midpoint of those two points. The midpoint of points P and Q can be calculated by doing the following:

(1+7/2),(11+7/2)=(4,9)

We already know that the line passes through the origin. So now we can calculate the slope since we have two points.

9-0/4-0=9/4 or 2.25
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 229
Joined: Tue Jan 13, 2009 6:56 am
Thanked: 8 times
GMAT Score:700

Re: slope of the line

by Uri » Sun Mar 01, 2009 6:24 am
rahulg83 wrote: My doubt is if the line is equidistant from the mentioned points, does that means mid-point of the line segment joining the two points P and Q lies on the line passing through origin.
yes
Top
User avatar
Legendary Member
Posts: 2469
Joined: Thu Apr 20, 2006 12:09 pm
Location: BtG Underground
Thanked: 85 times
Followed by:14 members

Re: slope of the line

by aim-wsc » Sun Mar 01, 2009 6:53 am
rahulg83 wrote:
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)?


OA later
Are you sure you typed question correctly, verbatim?
Because I think the meaning of the question changed...
If yes then, I agree with truplayer256

#your doubt regarding equidistant: Basically any point on the line, which is perpendicular to segment PQ is equidistant to P & Q. (that makes question ambiguous... )
Top
Senior | Next Rank: 100 Posts
Posts: 46
Joined: Sat Feb 21, 2009 8:02 am
Thanked: 1 times

by deepak_free » Sun Mar 01, 2009 10:07 am
Suppose the equation of line is ax+by+c=0,since line passes from origin, c=0 and the equation can be mx - y =0 or ax + by +c =0
where a = m,b=-1 and c=0, note here the slope is m
distance of a point(x1,y1) from a line ax+by+c=0 is
d = |(ax1+by1+c)|/sqrt(a^2+b^2) since the distance is equal from the points (1,11) and (7,7) the euqation to solve is
|m-11|=|7m+7|
which gives m = 2.25 hence answer is B.
Top
User avatar
Legendary Member
Posts: 575
Joined: Tue Nov 04, 2008 2:58 am
Location: India
Thanked: 18 times
Followed by:4 members
GMAT Score:710

Re: slope of the line

by rahulg83 » Sun Mar 01, 2009 7:29 pm
aim-wsc wrote:
rahulg83 wrote:
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)?


OA later
Are you sure you typed question correctly, verbatim?
Because I think the meaning of the question changed...
If yes then, I agree with truplayer256

#your doubt regarding equidistant: Basically any point on the line, which is perpendicular to segment PQ is equidistant to P & Q. (that makes question ambiguous... )
Actually what the question says is the line is equidistant from P and Q. That means the distance of line from P and that from Q is equal. This from nowhere means that mid-point of PQ lies on the given line. And the solution is given assuming this condition itself. Though deepak_free's solution is justified i feel...

P.S. Correct me please if i am wrong regarding the concept..
The answer indeed in 2.25..
Top
Master | Next Rank: 500 Posts
Posts: 221
Joined: Wed Jan 21, 2009 10:33 am
Thanked: 12 times
Followed by:1 members

by krisraam » Sun Mar 01, 2009 8:26 pm
what defines a line equidistant from 2 points?? Does that mean each and every point on the line is equidistant to P and Q. In that case the line has to be perpendicular to the line joining P and Q. The line thus formed will not go through the origin(slope = 2/3 one point is the mid point of P and Q (4,9)).

If the question says any one point on the line is equidistant to P and Q and that line passes through origin. We can have n number of lines satisfying that condition with different slopes. One possible line is the one that passes through the midpoint of P and Q.

This Question is confusing...

Thanks
raama
Top
Senior | Next Rank: 100 Posts
Posts: 42
Joined: Thu Feb 19, 2009 10:26 am
Thanked: 3 times
GMAT Score:720

by mygmat.2009 » Sun Mar 01, 2009 9:06 pm
IMO

Distance from a point to a line is defined as the length of the segment that lies on another line which passes through that point and is perpendicular to the first line.

In this case, any line passing through the midpoint of PQ would be equidistant from P and Q and since the question asks for the one that passes through the origin, then we know its equation will be of the form:
y = ax and midpoint (4,9) belongs to the line

==> Slope a = y/x = 9/4 = 2.25


Cheers,
Sam
Top
Master | Next Rank: 500 Posts
Posts: 221
Joined: Wed Jan 21, 2009 10:33 am
Thanked: 12 times
Followed by:1 members

by krisraam » Sun Mar 01, 2009 9:38 pm
mygmat.2009 wrote:IMO

Distance from a point to a line is defined as the length of the segment that lies on another line which passes through that point and is perpendicular to the first line.

In this case, any line passing through the midpoint of PQ would be equidistant from P and Q and since the question asks for the one that passes through the origin, then we know its equation will be of the form:
y = ax and midpoint (4,9) belongs to the line

==> Slope a = y/x = 9/4 = 2.25


Cheers,
Sam
The line formed is not perpendicular to PQ if it passes thru origin and midpoint of P and Q.

Thanks
raama
Top
Master | Next Rank: 500 Posts
Posts: 137
Joined: Thu Jan 08, 2009 1:27 am
Thanked: 7 times

by welcome » Mon Mar 02, 2009 1:19 am
Question is right as I have faced this Q in my test and got it right :).

Here is the answer..

1. say line is y=mx+c, going through origin so (0,0) will satisfy the eq. so c=0 hence eq is y=mx.

Now a point equidistent from (1, 11) and (7, 7) will be ( (1+7)/2, (11+7)/2 )

so this point will satisfy y=mx
=> 9=4m => m= 2.25
Answer B.
Shubham.
590 >> 630 >> 640 >> 610 >> 600 >> 640 >> 590 >> 640 >> 590 >> 590
Top
Senior | Next Rank: 100 Posts
Posts: 42
Joined: Thu Feb 19, 2009 10:26 am
Thanked: 3 times
GMAT Score:720

by mygmat.2009 » Mon Mar 02, 2009 6:31 am
krisraam, the line we are looking for is not perpendicular to PQ, I was referring to the line we draw to calculate the distance from a point to a line.

So, if point M is the mid-point of PQ, and O is the origin, then segment OM is not perpendicular to PQ, however, shortest distance from P to OM is the length of segment PP', where P' belongs to line OM and PP' perpendicular to OM. Same for Q, distance from Q to OM is QQ' , where Q' belongs to line OM and QQ' perpendicular to OM. Since line OM bisects PQ, then QQ' = PP'.

Do you agree?

Cheers,
Sam
Top
Post new topic Post Reply
• Page 1 of 1