Line problem

This topic has expert replies
Legendary Member
Posts: 1169
Joined: Sun Jul 06, 2008 2:34 am
Thanked: 25 times
Followed by:1 members

Line problem

by aj5105 » Tue Jan 13, 2009 9:28 am
If line L passes through point (m,n) and (-m,-n), where m and n are not 0, which of the following must be true?

I. The slope of L is positive.
II. The slope of L is negative.
III. L exactly passes through 2 quadrants.
Source: — Problem Solving |

Legendary Member
Posts: 527
Joined: Mon Jun 02, 2008 9:14 am
Location: Atlanta
Thanked: 17 times

by pandeyvineet24 » Tue Jan 13, 2009 9:37 am
Lets take point (m,n) as (1,1) -m,-n will be (-1,-1).
Slope is positive in this case.

Lets take point (m,n) as (-1,2) -m,-n will be (1,-2).
Slope of the line is negative in this case

III. L exactly passes through 2 quadrants. seems to be the best choice.

Y intercept of such a line will always be 0 and line will always pass through the origin. (However question says m and n cannot be 0).

User avatar
Legendary Member
Posts: 2134
Joined: Mon Oct 20, 2008 11:26 pm
Thanked: 237 times
Followed by:25 members
GMAT Score:730

by logitech » Tue Jan 13, 2009 9:45 am
pandeyvineet24 wrote:Lets take point (m,n) as (1,1) -m,-n will be (-1,-1).
Slope is positive in this case.

Lets take point (m,n) as (-1,2) -m,-n will be (1,-2).
Slope of the line is negative in this case

III. L exactly passes through 2 quadrants. seems to be the best choice.

Y intercept of such a line will always be 0 and line will always pass through the origin. (However question says m and n cannot be 0).
Okay,

so our equation y=mx+b satisfies both m,n and -m,-n

n = mx+b
-n = -mx +b

if we add these two equations we get

2b=0

and b=0

SO in other words, y=mx ( m can be negative or positive )

So the question stem has a flaw saying that m,n can not be ZERO.
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

User avatar
Legendary Member
Posts: 2134
Joined: Mon Oct 20, 2008 11:26 pm
Thanked: 237 times
Followed by:25 members
GMAT Score:730

by logitech » Tue Jan 13, 2009 9:47 am
but maybe the question wants to tell that M and N are not 0 in -m,-n and m,n pairs SO..

Y=mx can be still valid.

And III option does not need to be TRUE

I would choose NONE!
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

Legendary Member
Posts: 1169
Joined: Sun Jul 06, 2008 2:34 am
Thanked: 25 times
Followed by:1 members

by aj5105 » Tue Jan 13, 2009 10:05 am
answer is only III

use the formula y - y1 = m(x - x1)
where m = (y2 - y1)/(x2 - x1)

Substitute points.Substitute x=0,y=0 we get LHS = RHS which indicates that line passes through two quadrants.

Legendary Member
Posts: 527
Joined: Mon Jun 02, 2008 9:14 am
Location: Atlanta
Thanked: 17 times

by pandeyvineet24 » Tue Jan 13, 2009 10:14 am
i agree with you logitech,

but would not y = mx will always pass thru exactly 2 quadrants only

User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

Re: Line problem

by Stuart@KaplanGMAT » Tue Jan 13, 2009 10:47 am
aj5105 wrote:If line L passes through point (m,n) and (-m,-n), where m and n are not 0, which of the following must be true?

I. The slope of L is positive.
II. The slope of L is negative.
III. L exactly passes through 2 quadrants.
Let's back away from formulas and do some good ol' fashioned thinkin!

We don't know the signs of m and n. If we don't know the signs, then there's no way to determine whether the slope is positive or negative. Therefore, I and II are both out.

Our next step would depend on the answer choices. Please post the choices in the future!

If there's no choice that says "none of the above", then we'd pick the choice that says "III only" and wouldn't have to even test it.

If there's no choice that says "III only", then we'd pick the choice that says "none of the above" and wouldn't have to even test it.

If there are choices that say "III only" and "none of the above", then we have to test statement III - booooo!

So, how do we test (III)? Well, we could use some of them thar fancy arithmetic stuff or we could draw pretty pictures. The world needs more art, so I'd opt for drawing.

If we draw points (m,n) and (-m,-n) and connect them with a line, we'll quickly see that no matter which quadrants we pick for our two points, the line will always pass through the origin. Any line that passes through the origin only passes through 2 quadrants, so (III) must be true: choose III only.
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course

User avatar
Legendary Member
Posts: 2134
Joined: Mon Oct 20, 2008 11:26 pm
Thanked: 237 times
Followed by:25 members
GMAT Score:730

by logitech » Tue Jan 13, 2009 4:16 pm
III. L exactly passes through 2 quadrants.
:) Not the SECOND QUADRANTS!!!

I should write this on a black board for 1000 times :D
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

User avatar
Legendary Member
Posts: 2134
Joined: Mon Oct 20, 2008 11:26 pm
Thanked: 237 times
Followed by:25 members
GMAT Score:730

by logitech » Tue Jan 13, 2009 4:17 pm
pandeyvineet24 wrote:i agree with you logitech,

but would not y = mx will always pass thru exactly 2 quadrants only
True..I misread it. 2 and 2nd are different things! DUH :)
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

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

by welcome » Tue Jan 13, 2009 4:25 pm
logitech,
Your calculation is right but conclusion is wrong.
in y=mx+c, c indicates the constant which line is cutting on one of the axis. it is nothing to do with slop.

so y=mx is a equation of a line with no constant segment, but this line will pass through origin(0,0), as there points are satisfying y=mx. That means, y=mx will pass through 2 quadrants, Answer should be III.
Shubham.
590 >> 630 >> 640 >> 610 >> 600 >> 640 >> 590 >> 640 >> 590 >> 590

User avatar
Legendary Member
Posts: 2134
Joined: Mon Oct 20, 2008 11:26 pm
Thanked: 237 times
Followed by:25 members
GMAT Score:730

by logitech » Tue Jan 13, 2009 4:52 pm
welcome wrote:logitech,
Your calculation is right but conclusion is wrong.
in y=mx+c, c indicates the constant which line is cutting on one of the axis. it is nothing to do with slop.

so y=mx is a equation of a line with no constant segment, but this line will pass through origin(0,0), as there points are satisfying y=mx. That means, y=mx will pass through 2 quadrants, Answer should be III.
My conclusion is not wrong.

I found out that y=mx+0 and per question M and N are not zero , which means that slope can not be 0 or undefined.

I thought that statement III is asking whether the line MUST pass through the 2nd quadrant; ( it may or it may not) , but the line will definitely cross through either I and III quadrants or II and IV quadrants.

So this piece of information makes the III statement valid.
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"