XY Coordinate

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relaxin99: Senior | Next Rank: 100 Posts; Posts: 73; Joined: Tue Sep 16, 2008 4:33 pm; Thanked: 2 times

by relaxin99 » Thu Feb 12, 2009 2:36 pm

In the xy coordinate plane, line L and line K intersect at point (4, 3). Is the product of their slops negative?

(1) The product of the x-intercepts of the line L and K is positive
(2) The product of the lines L and K are negative

The answer is C, can someone please explain

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awesomeusername: Master | Next Rank: 500 Posts; Posts: 226; Joined: Tue Jan 13, 2009 1:27 pmFollowed by: 1 member; Thanked: 23 times

by awesomeusername » Thu Feb 12, 2009 5:15 pm

I think you mean

In the XY-coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?

1) The product of the x-intercepts of line L and K is positive.

2) The product of the y-intercepts of line L and k is negative.

We know that the lines L and K intersect in Quadrant I (the top right quadrant).

The products of the slopes can be negative if the slopes are opposite signs. For examples, L = positive slope, K = negative slope, and vice versa.

1) The product of the x-intercepts being positive means that they both cross the x-axis BOTH where x < 0 , or where x > 0. But this does not imply that their slopes have to be the opposite signs of each other.

2) The product of the Y intercepts being negative means that one line crosses the y-axis above the line y=0, and the other line crosses the y-axis below the line where y=0. BUT, this does not imply that they have opposite sign slopes.

If both conditions were true, then the slopes MUST be opposite signs of each other.

Attachments

: Showing that Case 1 doesn't imply that the product of their slopes are negative

: Showing that Case 2 doesn't imply that the products of the slopes are negative

: Showing that if BOTH Case 1 and Case 2 are true, then the slopes MUST be the opposite. Thus, the products are negative.

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logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

Re: XY Coordinate

by logitech » Thu Feb 12, 2009 6:22 pm

K=a1x+b1
L=a2x+b2

is a1a2 < 0 ?

St1)

(-b1/a1) x (- b2/a2) > 0

or b1b2/a1a2 > 0

We can't decide whether a1a2 < 0 because we need to know the sign of b1b2

INSUF

St2)

b1b2 < 0 Nothing given for a1a2

INSUF

St1+St2)

b1b2/a1a2 > 0
b1b2 < 0

Hence, a1a2 < 0

Choose (C)

LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

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Bidisha800: Master | Next Rank: 500 Posts; Posts: 344; Joined: Sun Sep 28, 2008 11:00 amFollowed by: 1 member; Location: USA; Thanked: 6 times

by Bidisha800 » Thu Feb 12, 2009 7:57 pm

A straight line can be expreesed as :

x/a1+y/b1=1 (a1= X axis intercept, b1=Y axis intercept)

b1x+a1y=a1b1

y=-(b1/a1)x+a1b1 .......line K

similarly y=-(b2/a2)x+a2b2 .....line L

product of slopes = (b1b2/a1a2)

From (1) a1a2 >0 INSUFF
From (2) b1b2 >0 INSUFF

Combining product of slopes is positive

(C)

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logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

by logitech » Fri Feb 13, 2009 2:23 am

Bidisha800 wrote:A straight line can be expreesed as :

x/a1+y/b1=1 (a1= X axis intercept, b1=Y axis intercept)

b1x+a1y=a1b1

y=-(b1/a1)x+a1b1 .......line K

similarly y=-(b2/a2)x+a2b2 .....line L

product of slopes = (b1b2/a1a2)

From (1) a1a2 >0 INSUFF
From (2) b1b2 >0 INSUFF

Combining product of slopes is positive

(C)

Bringing tears to my eyes

LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

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Bidisha800: Master | Next Rank: 500 Posts; Posts: 344; Joined: Sun Sep 28, 2008 11:00 amFollowed by: 1 member; Location: USA; Thanked: 6 times

by Bidisha800 » Sat Feb 14, 2009 10:30 am

logitech wrote:
Bidisha800 wrote:A straight line can be expreesed as :

x/a1+y/b1=1 (a1= X axis intercept, b1=Y axis intercept)

b1x+a1y=a1b1

y=-(b1/a1)x+a1b1 .......line K

similarly y=-(b2/a2)x+a2b2 .....line L

product of slopes = (b1b2/a1a2)

From (1) a1a2 >0 INSUFF
From (2) b1b2 >0 INSUFF

Combining product of slopes is positive

(C)
Bringing tears to my eyes

Here is a suggestion JACKA$$ - Wipe your last 3 sub 580 scores with your tears then rent a gun, buy a bullet and kill your self :twisted:

Drill baby drill !

GMATPowerPrep Test1= 740
GMATPowerPrep Test2= 760
Kaplan Diagnostic Test= 700
Kaplan Test1=600
Kalplan Test2=670
Kalplan Test3=570

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by beatthegmat » Sun Feb 15, 2009 3:08 pm

I can't accept language like that here.

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FC: Senior | Next Rank: 100 Posts; Posts: 37; Joined: Wed Feb 18, 2009 5:13 pm; Thanked: 3 times; GMAT Score:690

by FC » Thu Jul 30, 2009 7:25 am

Am wrong or the "point (4,3)" information is just irrelevant ?

From (I) (bl*bx)/(ml*mk)>0 (INSUF), from (II) bk*bl<0 (INSUF), so with both (I) and (II) together you can answer if ml*mk<0. So the info about (4,3) is not needed.

Thanks,

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Pdgmat2010: Master | Next Rank: 500 Posts; Posts: 176; Joined: Sun Feb 21, 2010 6:00 am; Thanked: 2 times; GMAT Score:710

by Pdgmat2010 » Fri Aug 20, 2010 3:13 am

yes, if you look at it from a purely 'equation of a line perspective' the point (4,3) info is not needed
product of slopes ( refer logitech's calculation) = b1b2/a1a2

to check if b1b2/a1a2 is < 0 we just need to know the signs of the numerator and the denominator and we are through!
Stmt 1 does just that but only provides sign of the denominator. so insufficient a1a2 >0

Stmt 2: provides sign of the numerator. numerator b1b2 < 0

only after combining the 2 statements can we see that b1b2/a1a2 < 0.
and that answers our question.

Drawing out each and every possibility on the graph is a tedious process.
I find the mathematical equivalent quite quick , if not simple.

a few general pointers
1. any linear equation of a line can be expressed as ax+by+c=0
ax+by = c
y-intercept is value of y when x =0 , i.e. y int= -c/b
x-intercept is value of x when y=0 , i.e. x int = -c/a
slope of a line passing through two points ( 0,-c/b) and (-c/a,0) is
(-c/b-0)/(0-(-c/a)) = - b/a
slope of a line = negative of y intercept / x intercept
its easier to remember this formula rather than its derivation.

FC wrote:Am wrong or the "point (4,3)" information is just irrelevant ?

From (I) (bl*bx)/(ml*mk)>0 (INSUF), from (II) bk*bl<0 (INSUF), so with both (I) and (II) together you can answer if ml*mk<0. So the info about (4,3) is not needed.

Thanks,

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AnujVasa: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Wed Sep 21, 2011 6:09 am

by AnujVasa » Sun Nov 27, 2011 8:44 am

Hey!

(-c/b-0)/(0-(-c/a)) = - b/a

Should'nt the result of division be -a/b as -c/b*a/c=-a/b.

Appreciate your help!!!!

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venmic: Master | Next Rank: 500 Posts; Posts: 222; Joined: Mon Oct 13, 2008 4:04 pmFollowed by: 2 members; Thanked: 3 times

by venmic » Tue Oct 09, 2012 1:08 pm

Bidisha800 ........... you are the meanest guy I have come acroos on any of the forums.. people like you should be banned from site for demeaning guys who are learning and looking for encoraugement .. shame....

Bidisha800 wrote:
logitech wrote:
Bidisha800 wrote:A straight line can be expreesed as :

x/a1+y/b1=1 (a1= X axis intercept, b1=Y axis intercept)

b1x+a1y=a1b1

y=-(b1/a1)x+a1b1 .......line K

similarly y=-(b2/a2)x+a2b2 .....line L

product of slopes = (b1b2/a1a2)

From (1) a1a2 >0 INSUFF
From (2) b1b2 >0 INSUFF

Combining product of slopes is positive

(C)
Bringing tears to my eyes
Here is a suggestion JACKA$$ - Wipe your last 3 sub 580 scores with your tears then rent a gun, buy a bullet and kill your self :twisted:

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