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Slope

by Vemuri » Wed Mar 18, 2009 6:50 pm
Are lines p (with slope m) and q (with slope n) perpendicular to each other?

1. m + 2 = n
2. m + n = 0

I don't know the answer to this question. [spoiler]I feel both the statements are required to answer the question (when combined the values of m & n are -1 & 1 respectively, which satisfies the perpendicular slope concept). Any contentions?[/spoiler]
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by DanaJ » Wed Mar 18, 2009 10:39 pm
You actually solved this one correctly: two lines are perpendicular if the product of their slopes is -1. As you've pointed out, only by using both stmts can you reach that conclusion.
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by kaf » Thu Mar 19, 2009 4:06 am
DanaJ wrote:You actually solved this one correctly: two lines are perpendicular if the product of their slopes is -1. As you've pointed out, only by using both stmts can you reach that conclusion.

Could you please explain how the products of their slopes is -1

thanks
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by sureshbala » Thu Mar 19, 2009 8:52 am
kaf wrote:
DanaJ wrote:You actually solved this one correctly: two lines are perpendicular if the product of their slopes is -1. As you've pointed out, only by using both stmts can you reach that conclusion.

Could you please explain how the products of their slopes is -1

thanks
Slope of a line is given by m = tanθ, where θ is the angle made by the line in the positive direction (anti-clock wise) of the X axis. So imagine that two perpendicular lines are in the first quadrant. If one of them makes an angle θ in the positive direction of X axis the other perpendicular line makes an angle 90+θ.

Hence product of their slopes = tanθ x tan(90+θ) = tanθ x -Cotθ = -1.

You can extend this to any quadrants....
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by gmat740 » Thu Mar 19, 2009 11:34 am
https://www.beatthegmat.com/explanation- ... 33100.html


check out the last reply of this thread.
I have explained it over there
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by PinkBox » Thu Mar 19, 2009 2:42 pm
use property mentioned by others above: the product of the slope of two perpendicular lines=-1
statement I
m+2=n
(m+2)m=-1
m^2+2m+1=0
m=-1, n=1
this would mean that is it possible for the two lines to be perpendicular but m and n could take other values to make the statement false. so insufficient
statement II
m=-n
-n^2=-1
n= plus or minus 1
m=plus or minus 1
same idea. the two lines could be perpendicular but they do not have to be.
combine statements 1 and 2
if n=1 and m=-1 the two lines are perpendicular, otherwise they are not.
hence E
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Re: Slope

by El Cucu » Thu Mar 19, 2009 3:26 pm
Vemuri wrote:Are lines p (with slope m) and q (with slope n) perpendicular to each other?

1. m + 2 = n
2. m + n = 0

I don't know the answer to this question. [spoiler]I feel both the statements are required to answer the question (when combined the values of m & n are -1 & 1 respectively, which satisfies the perpendicular slope concept). Any contentions?[/spoiler]
Perpendicular slope= - 1/m

1. m=n+2 If m=-1 n=1 Perpendicular. If m=1 n=-1 Perpendicular. If m=2 n=4 not perpendicular.

2.m=-n If m=1 n=-1 Perpendicular If m=-1, n=1 Perfpendicular But If m=3 n=-3 If m=6 n=-6 not perpendicular.

IMO E).
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Re: Slope

by Stuart@KaplanGMAT » Thu Mar 19, 2009 3:31 pm
Vemuri wrote:Are lines p (with slope m) and q (with slope n) perpendicular to each other?

1. m + 2 = n
2. m + n = 0

I don't know the answer to this question.
Two lines are perpendicular if their slopes are negative reciprocals of one another.

(1) m + 2 = n

Could have m=-1 and n=1 to get a "yes".
Could have m=12 and n = 14 to get a "no".

Insufficient.

(2) m + n = 0

could have m = -1 and n = 1 to get a "yes".

Could have m = -2 and n = 2 to get a "no".

Insufficient.

Together: two equations, two unknowns; therefore, we can solve for m and n. If we can solve for both variables, we can answer ANY question about the two variables: sufficient.

Apart insufficient, Together sufficient: choose (C).
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