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# M03-15

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Math Expert
Joined: 02 Sep 2009
Posts: 49968

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16 Sep 2014, 00:20
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Difficulty:

15% (low)

Question Stats:

73% (00:18) correct 27% (00:41) wrong based on 159 sessions

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If two lines have slopes $$m$$ and $$n$$, respectively, are they perpendicular?

(1) $$m * n = -1$$

(2) $$m = - n$$

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Joined: 02 Sep 2009
Posts: 49968

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16 Sep 2014, 00:20
Official Solution:

For one line to be perpendicular to another, the relationship between their slopes has to be negative reciprocal, so if the slope of one line is $$m$$ then the line perpendicular to it will have the slope $$-\frac{1}{m}$$. In other words, the two lines are perpendicular if and only if the product of their slopes is -1.

(1) $$m*n = -1$$. Directly gives an answer YES to the question.

(2) $$m = -n$$. Now, if for example $$m=3=-(-3)=-n$$ then the lines are not perpendicular but if $$m=1=-(-1)=-n$$, so if $$m=1$$ and $$n=-1$$ (or $$m=-1$$ and $$n=1$$) then as $$mn$$ would be equal to -1 the lines would be perpendicular. Not sufficient.

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Joined: 21 Mar 2017
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19 Jul 2017, 06:56
Hi Bunuel,
Could you please elaborate a little on Bold part.
(2) m=−nm=−n. Now, if for example m=3=−(−3)=−nm=3=−(−3)=−n then the lines are not perpendicular but if m=1=−(−1)=−nm=1=−(−1)=−n,

So do you mean to say in the second case when we put the value in the reciprocal equation that is essential for perpendicular condition m*n=3*(-3)=-9 which is not the reciprocal i.e equal to -1. But in the case of 1 and -1 it is.
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Yet at the hundred and first blow it would split in two.
And I knew it was not that blow that did it, But all that had gone Before
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Joined: 02 Sep 2009
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19 Jul 2017, 08:35
1
Prashant10692 wrote:
Hi Bunuel,
Could you please elaborate a little on Bold part.
(2) m=−nm=−n. Now, if for example m=3=−(−3)=−nm=3=−(−3)=−n then the lines are not perpendicular but if m=1=−(−1)=−nm=1=−(−1)=−n,

So do you mean to say in the second case when we put the value in the reciprocal equation that is essential for perpendicular condition m*n=3*(-3)=-9 which is not the reciprocal i.e equal to -1. But in the case of 1 and -1 it is.

The two lines are perpendicular if and only if the product of their slopes is -1.

If $$m=3$$ and $$n=-3$$, then the product mn is not -1, so the lines are not perpendicular.
If $$m=1$$ and $$n=-1$$, then the product mn is -1, so the lines are perpendicular.

Two different answers, hence not sufficient.
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29 Sep 2018, 12:36
Bunuel wrote:
Official Solution:

For one line to be perpendicular to another, the relationship between their slopes has to be negative reciprocal, so if the slope of one line is $$m$$ then the line perpendicular to it will have the slope $$-\frac{1}{m}$$. In other words, the two lines are perpendicular if and only if the product of their slopes is -1.

(1) $$m*n = -1$$. Directly gives an answer YES to the question.

(2) $$m = -n$$. Now, if for example $$m=3=-(-3)=-n$$ then the lines are not perpendicular but if $$m=1=-(-1)=-n$$, so if $$m=1$$ and $$n=-1$$ (or $$m=-1$$ and $$n=1$$) then as $$mn$$ would be equal to -1 the lines would be perpendicular. Not sufficient.

Hi Bunuel,

Regarding statement 1, is it not possible for M to be 1 and N -1?

This was the conflict I had in this question. I felt it was a trick answer.
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Joined: 02 Sep 2009
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30 Sep 2018, 04:30
Mfon wrote:
Bunuel wrote:
Official Solution:

For one line to be perpendicular to another, the relationship between their slopes has to be negative reciprocal, so if the slope of one line is $$m$$ then the line perpendicular to it will have the slope $$-\frac{1}{m}$$. In other words, the two lines are perpendicular if and only if the product of their slopes is -1.

(1) $$m*n = -1$$. Directly gives an answer YES to the question.

(2) $$m = -n$$. Now, if for example $$m=3=-(-3)=-n$$ then the lines are not perpendicular but if $$m=1=-(-1)=-n$$, so if $$m=1$$ and $$n=-1$$ (or $$m=-1$$ and $$n=1$$) then as $$mn$$ would be equal to -1 the lines would be perpendicular. Not sufficient.

Hi Bunuel,

Regarding statement 1, is it not possible for M to be 1 and N -1?

This was the conflict I had in this question. I felt it was a trick answer.

Even in this case the slope of one line is a negative reciprocal of the slope of another line.
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Re: M03-15 &nbs [#permalink] 30 Sep 2018, 04:30
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# M03-15

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