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16 Sep 2014, 00:20
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If two lines have slopes \(m\) and \(n\) respectively, are they perpendicular?
(1) \(m * n = -1\)
(2) \(m = - n\)
(1) \(m * n = -1\)
(2) \(m = - n\)
16 Sep 2014, 00:20
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Official Solution:
If two lines have slopes \(m\) and \(n\) respectively, are they perpendicular?
To determine if one line is perpendicular to another, we need to examine the relationship between their slopes. Two lines are perpendicular if their slopes are negative reciprocals of each other. This means if one line has a slope of \(m\), the line perpendicular to it should have a slope of \(-\frac{1}{m}\). Thus, the two lines will be perpendicular if and only if the product of their slopes is -1.
(1) \(m*n = -1\).
This directly answers the question with a YES.
(2) \(m = -n\).
If \(m=3\) and \(n=-3\), the lines are not perpendicular. However, if \(m=1\) and \(n=-1\) (or \(m=-1\) and \(n=1\)), the product \(m*n = -1\), confirming that the lines are perpendicular. Not sufficient.
Answer: A
If two lines have slopes \(m\) and \(n\) respectively, are they perpendicular?
To determine if one line is perpendicular to another, we need to examine the relationship between their slopes. Two lines are perpendicular if their slopes are negative reciprocals of each other. This means if one line has a slope of \(m\), the line perpendicular to it should have a slope of \(-\frac{1}{m}\). Thus, the two lines will be perpendicular if and only if the product of their slopes is -1.
(1) \(m*n = -1\).
This directly answers the question with a YES.
(2) \(m = -n\).
If \(m=3\) and \(n=-3\), the lines are not perpendicular. However, if \(m=1\) and \(n=-1\) (or \(m=-1\) and \(n=1\)), the product \(m*n = -1\), confirming that the lines are perpendicular. Not sufficient.
Answer: A
19 Jul 2017, 08:35
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Prashant10692
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.
