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A line l in the xy-coordinate system passes through the origin and poi 27 Jul 2020, 10:00
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

25% (medium)

Question Stats:

69% (01:30) correct 31% (01:43) wrong based on 48 sessions

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A line \(l\) in the \(xy\)-coordinate system passes through the origin and point (\(m,n\)). Is \(m<n\)?
1) The line \(l\) has a negative slope
2) \(m<0\)
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C
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Siak0730
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Schools: ISB '25 (A)
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A line l in the xy-coordinate system passes through the origin and poi 27 Jul 2020, 11:34
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TheUltimateWinner
A line \(l\) in the \(xy\)-coordinate system passes through the origin and point (\(m,n\)). Is \(m<n\)?
1) The line \(l\) has a negative slope
2) \(m<0\)
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line equation : \(y= Mx+c\)

slope of a line passing through two points : \(M = (y2-y1)/(x2-x1)\)
well, don't have to memorize it, since line passes through two points, the basic equation must satisfy these two points.
lets say two points be \(A: (x1, y1)\) and \(B: (x2, y2)\)
let line C passes through A and B
line C standard equation: \(y= Mx+c\)

now, since line C passes through \(A\) \((x1, y1)\) we have
\(y1 = Mx1+c \)
\(y1-Mx1 = c\)

since line C passes through \(B (x2, y2)\), we have
\(y2 = Mx2+c\)
from above we substitute c
\(y2= Mx2+y1-Mx1\)
or
\(y2-y1 = M(x2-x1)\)
\(M= (y2-y1)/(x2-x1)\\
\)

similarly, we can workout in this question too
point 1: origin (0,0)
\(0 = 0+c => c=0\)

point 2: (m,n)
\(n=Mm+c\)
or \(M= (n-c)/m\)
from point 1 we get c=0
\(M= n/m\)

QUESTION: \(m < n?\)

STATEMENT 1 : slope M is negative
\(M = n/m\)
\(M < 0\)
either n can be \(< 0\)
or m can be \(< 0\)

therefore, NOT SUFFICIENT

STATEMENT 2 : \(m<0\)
no information about n
what if n < m < 0

therefore, NOT SUFFICIENT

combining STATEMENT 1 and STATEMENT 2, we get
\(M < 0\) or \(n/m <0\)
and \(m < 0\)
which implies
\(n > 0 >m\)
and hence, \(n > m\)

therefore, SUFFICIENT and C


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