Does line k intersect quad II?

[denise.lu]

DS question - does line k intersect quad II (upper left hand quadrant)?
1) slope of K = -1/6
2) y-int of k = -6

The answer is A but not quite sure to get this?

[RonPurewal]

DS question - does line k intersect quad II (upper left hand quadrant)?
1) slope of K = -1/6
2) y-int of k = -6

The answer is A but not quite sure to get this?

the 1/6 is irrelevant; all that matters in (1) is that the slope is negative.
lines with negative slopes go up to the left, down to the right. this means that, if you follow ANY negatively sloped line far enough to the left, it will go up into the second quadrant.
negatively sloped lines MUST hit quadrants 2 and 4.
positively sloped lines MUST hit quadrants 1 and 3.
zero exceptions.
the other quadrants are mutually exclusive options. i.e., if a positively sloped line hits quadrant 2, then it can't hit quadrant 4, and vice versa. same thing with quadrants 1 and 3 for negatively sloped lines.

with statement (2), a horizontal or positively sloped line through (0, -6) won't hit quadrant 2, but a negatively sloped line will, so that's insufficient.

[denise.lu]

Thanks Ron!

[JonathanSchneider]

: )

[nitin_prakash_khanna]

Instructors,
is this really true for a negative slope line to hit the 2nd and 4th quadrant as Ron explained.

Lets say i have a line x=3 which is parallel to y axis. Now if i just turn it to the left, just a tiny bit may be by few seconds (less than a degree). So that the slope is negative and if i extend this line from negative infinity to positive infinity, will it neccessary hit the 2nd quadrant? Can it just stay in 1st and 4th quadrant like the the x=3 line.?

Thanks in advance for a response.

[anoo.anand]

yes of course it will pass thru 2nd and 4th quadrant unless it is finite lenth line...

[nitin_prakash_khanna]

Thanks Understand it.
If the lines are not parallel somewhere they will intersect.

[RonPurewal]

So that the slope is negative and if i extend this line from negative infinity to positive infinity, will it neccessary hit the 2nd quadrant?

yes.

Can it just stay in 1st and 4th quadrant like the the x=3 line.?

no.

let's say the slope is -0.000000001 (i.e., negative one-billionth).
still, this means that it moves to the left one unit for every billion that it goes up.
move over enough billions, and you'll make it.
apply the same reasoning to any other negative slope.

[nitin_prakash_khanna]

Yes it will have to go a lots of billions up there but will go to 2nd quadrant for sure.
Thanks for your explanation Ron, appreciate it.

[Ben Ku]

Glad it helped.

[herogmat]

Hi Ron,
I understand the concept of negative , positive , zero and undefined slopes. Thanks for your explanation. I have a doubt though. I am not questioning the official answer - no way :-)
You mentioned below:
QUOTE
negatively sloped lines MUST hit quadrants 2 and 4.
positively sloped lines MUST hit quadrants 1 and 3.
UNQUOTE
I am just wondering that if the negatively slope line is 'finite' and don't go upto the quardant 2 . Is there any specific reason that we should not at all think about that possibility ?

[ps63739]

If it mentions a line segment (within finite points) then we can think about that.. However, if it mentions just a line, there I don't think we should.

[herogmat]

I see ... You mean to say , if it is mentioned 'line' , then we should treat that as an infinite line. For this to be finite , it must be mentioned as line segment.
Thanks

[RonPurewal]

I see ... You mean to say , if it is mentioned 'line' , then we should treat that as an infinite line. For this to be finite , it must be mentioned as line segment.
Thanks

yes.

lines don't stop.