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Re: M23 #14 - Incorrect OA ? [#permalink]
02 Jan 2011, 05:11
The OA is D.
OE : For lines to be parallel, their slopes must be equal. The second equation can be rewritten as y = x/k - b/k . Because slopes must be equal, k=1/k or k^2=1 or |k| = 1.
But for k=-1 the 2 equations are effectively the same (i.e. Not parallel)
Re: M23 #14 - Incorrect OA ? [#permalink]
02 Jan 2011, 07:43
Yes, when you solve the problem there will be 2 possible solutions to k, -1 and 1. But look at what is being asked in the q, are these lines parallel. And for k=-1, these lines are essentially the same y = -x + b. Hence not parallel.
Re: M23 #14 - Incorrect OA ? [#permalink]
02 Jan 2011, 08:26
TheBirla wrote:
Yes, when you solve the problem there will be 2 possible solutions to k, -1 and 1. But look at what is being asked in the q, are these lines parallel. And for k=-1, these lines are essentially the same y = -x + b. Hence not parallel.
Posted from my mobile device
Yeah you are right, sorry. Maybe someone can shed some light on this ? _________________
Re: M23 #14 - Incorrect OA ? [#permalink]
02 Jan 2011, 08:46
D is the right answer
y=kx+b y=x/k - b/k
Since slopes are equal k = 1/k k^2=1 k=+1 or -1 Therefore |k| = 1 ---> |k|-1 = 0
If k=1 , we knw hw the equs would look like
y=x+b y=x-b
If k=-1 , then two equations would look like
y = -x + b y = -x +b
In both the scenarios the co-efficient of x is same , so the lines are parallel , the constant b can be +ve or -ve ....that does not matter . if a line is parallel , it does not mean that the constants must have the same sign . It depends on in which quadrant the lines are . A line passing through 1st and 4th quadrant can be parallel to line passing through 2nd and 3rd quadrant . In such scenarios , the slope of both the lines would be equal BUT NOT the constants...... _________________
Re: M23 #14 - Incorrect OA ? [#permalink]
02 Jan 2011, 08:49
ysr wrote:
Option B cannot be the answer as it has only +1 ....what about -1 ?
The question is whether a line can be both parallell and equal at the same time. It's obvious that if the lines are equal they have the same slope.
The common definition would be that two lines are parallell if they have the same slope and do not intersect. Two equal lines would intersect at every point, so could you really say that two identical lines are parallell? _________________
Re: M23 #14 - Incorrect OA ? [#permalink]
02 Jan 2011, 09:12
ysr wrote:
If k=-1 , then two equations would look like
y = -x + b y = -x +b
In both the scenarios the co-efficient of x is same , so the lines are parallel , the constant b can be +ve or -ve ....that does not matter . if a line is parallel , it does not mean that the constants must have the same sign . It depends on in which quadrant the lines are . A line passing through 1st and 4th quadrant can be parallel to line passing through 2nd and 3rd quadrant . In such scenarios , the slope of both the lines would be equal BUT NOT the constants......
I agree that the sign of constant does not matter, but the constant itself does. And as Mackieman has pointed out already, if the constant of 2 parallel lines is the same, they are not parallel by definition, i.e. they are identical lines, BUT not parallel.
Re: M23 #14 - Incorrect OA ? [#permalink]
30 Mar 2012, 01:16
TheBirla wrote:
The OA is D.
OE : For lines to be parallel, their slopes must be equal. The second equation can be rewritten as y = x/k - b/k . Because slopes must be equal, k=1/k or k^2=1 or |k| = 1.
But for k=-1 the 2 equations are effectively the same (i.e. Not parallel)
Hence i think Ans should be B.
I also have the same question. Can anyone put more light on it _________________