Lines a and b have different y-intercepts. When line a is reflected around the y-axis, is its reflection parallel to line b?
(1) Line a is perpendicular to line b.
(2) The slope of line b > 0.
Source: Grockit
OA: E
While solving this problem, I thought the answer is A, but it turns out to be E
Slopes of line
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To prove that it's (e):psm12se wrote:Lines a and b have different y-intercepts. When line a is reflected around the y-axis, is its reflection parallel to line b?
(1) Line a is perpendicular to line b.
(2) The slope of line b > 0.
Source: Grockit
OA: E
While solving this problem, I thought the answer is A, but it turns out to be E
* Draw line b with slope 1. Then draw line a perpendicular to it. Then do the reflection, and watch what happens.
* Draw line b with any positive slope in the world except 1. Then draw line a perpendicular to it. Then do the reflection, and watch what happens.
Give it a shot.
Ron has been teaching various standardized tests for 20 years.
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lunarpower wrote:To prove that it's (e):psm12se wrote:Lines a and b have different y-intercepts. When line a is reflected around the y-axis, is its reflection parallel to line b?
(1) Line a is perpendicular to line b.
(2) The slope of line b > 0.
Source: Grockit
OA: E
While solving this problem, I thought the answer is A, but it turns out to be E
* Draw line b with slope 1. Then draw line a perpendicular to it. Then do the reflection, and watch what happens.
* Draw line b with any positive slope in the world except 1. Then draw line a perpendicular to it. Then do the reflection, and watch what happens.
Give it a shot.
Hi RON,
Initially even I thought answer as A. But, later drawing the graph I came to know answer A is in-correct.
But, the actual problem is it consumes more time to draw the graph as given.
Since almost all the GMAT problems can be solved by more than one method is there any other way we would interpret the solution.
I thank you in advance.
Regards,
Uva.
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it shouldn't take more than ten or twenty seconds.Uva@90 wrote:Initially even I thought answer as A. But, later drawing the graph I came to know answer A is in-correct.
But, the actual problem is it consumes more time to draw the graph as given.
what are you drawing? why is it taking so long?
if you actually had to use a slope formula, then i can see how this might take a little longer. but, two things:
1/ it still shouldn't take longer than a minute or so,
2/ you should know what different slope values look like on a graph, without having to think about coordinate values. (for instance, a slope of +1 makes a 45º angle with the coordinate axes; this is something you should know as a thing, without having to think about ∆y/∆x.
with "reflection" you're going to have to draw it, unless you've memorized what happens.Since almost all the GMAT problems can be solved by more than one method is there any other way we would interpret the solution.
e.g., if you reflect a line with a slope of "m" across either axis, then you'll get a line with a slope of "-m".
i wouldn't be able to remember this fact -- i'd have to draw a picture and look at it -- but maybe you can.
in other words, the only "alternative" here would be to have advance knowledge of what happens to the slope when you reflect the line.
in general, i wouldn't recommend accumulating too much of that kind of memorized knowledge -- if you have too many facts bumping around in your head, it will become harder to develop any intuition about the situation.
Ron has been teaching various standardized tests for 20 years.
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by the way, the gmat contains A LOT of coordinate geometry problems on which you'll basically have to draw a picture of the situation. sometimes there will be workarounds, but (a) sometimes there won't, and (b) some of the workarounds are really, really hard and/or would require an extremely unusual amount of memorized info (as is the case here).
in other words... you should be able to draw coordinate-geometry situations quickly and easily. you should get to a point where making the drawing is not an issue, so that you can use your brainpower to analyze the actual situation rather than worrying about how to draw it.
this is especially the case in data sufficiency, on which very few coordinate problems are algebraic, and nearly all are highly visual.
in other words... you should be able to draw coordinate-geometry situations quickly and easily. you should get to a point where making the drawing is not an issue, so that you can use your brainpower to analyze the actual situation rather than worrying about how to draw it.
this is especially the case in data sufficiency, on which very few coordinate problems are algebraic, and nearly all are highly visual.
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- Uva@90
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Ron,
I took Some two points(without any slope thing).Since it makes the reflection around the Y axis. I made to change the sign of y value alone and redrawn it. That's why it took long time.
Any way Thanks Ron for your information.
I took Some two points(without any slope thing).Since it makes the reflection around the Y axis. I made to change the sign of y value alone and redrawn it. That's why it took long time.
Any way Thanks Ron for your information.
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Well, two things.Uva@90 wrote:Ron,
I took Some two points(without any slope thing).Since it makes the reflection around the Y axis. I made to change the sign of y value alone and redrawn it. That's why it took long time.
Any way Thanks Ron for your information.
1/
The y-coordinates don't change here; the x-coordinates do. If you change the y-coordinates, you are actually reflecting across the x-axis. (Try it with a random point.)
On the other hand, if you change "y axis" to "x axis" in the problem statement, then the correct answer (as well as the correct thought process) is still the same. So you can get away with this mistake here.
2/
If you realize that the sign of one coordinate changes, then you don't actually have to draw the situation. Just realize that slope = "rise"/"run".
If you change both x-coordinates, then "run" is the opposite of what it originally was. The "rise" doesn't change, so you the new slope is the opposite of the old one.
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Yes Ron you are right. When reflection is made across Y-axis then sign changes for x value.1/
The y-coordinates don't change here; the x-coordinates do. If you change the y-coordinates, you are actually reflecting across the x-axis. (Try it with a random point.)
On the other hand, if you change "y axis" to "x axis" in the problem statement, then the correct answer (as well as the correct thought process) is still the same. So you can get away with this mistake here.
That's a nice clue Ron. I will keep remember.2/
If you realize that the sign of one coordinate changes, then you don't actually have to draw the situation. Just realize that slope = "rise"/"run".
If you change both x-coordinates, then "run" is the opposite of what it originally was. The "rise" doesn't change, so you the new slope is the opposite of the old one.
I thank you once again for identifying my mistakes and helping me in understanding the concept.
Regards,
Uva.
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This post doesn't give me a sense of what you don't understand, so I can't give a specific response.bsahil wrote:I am still unable to visualise..can somebody please explain
But, do this:
* Put a few points on an x-y coordinate system.
* Reflect them across the y-axis (I assume you know what this means).
* Watch what happens to the coordinates.
* Generalize.
If you are unsure even of what "reflect across the y-axis" means, then take that exact phrase, type it into google, and read what you get.
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you're welcome. (and thank you, too, for actually asking specific questions, rather than just throwing a problem up there with no commentary.)Uva@90 wrote:I thank you once again for identifying my mistakes and helping me in understanding the concept.
Regards,
Uva.
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This way of explanation is most helpful for us. I wish I can see at least one explanation (like this one) of each question type.lunarpower wrote:
To prove that it's (e):
* Draw line b with slope 1. Then draw line a perpendicular to it. Then do the reflection, and watch what happens.
* Draw line b with any positive slope in the world except 1. Then draw line a perpendicular to it. Then do the reflection, and watch what happens.
Give it a shot.
Thanks RON.
lunarpower wrote:This post doesn't give me a sense of what you don't understand, so I can't give a specific response.bsahil wrote:I am still unable to visualise..can somebody please explain
But, do this:
* Put a few points on an x-y coordinate system.
* Reflect them across the y-axis (I assume you know what this means).
* Watch what happens to the coordinates.
* Generalize.
If you are unsure even of what "reflect across the y-axis" means, then take that exact phrase, type it into google, and read what you get.
I mean to say...when we reflect line a across y axis then the y co-ordinates of the line will change in sign. Since sign of y coordinate changes then slope of the line a is also multiplied with -1.
Then what happens...s it never possible that line a becomes parallel to line b.
May be my view is absolutely silly and i am missing something...
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^^ yes...bsahil wrote:I mean to say...when we reflect line a across y axis then the y co-ordinates of the line will change in sign. Since sign of y coordinate changes then slope of the line a is also multiplied with -1.
i don't understand your reasoning here, because, well, you haven't provided any of the reasoning here.Then what happens...s it never possible that line a becomes parallel to line b.
basically, here's the point:
* line b is perpendicular to line a. therefore, (slope of line b) = -1/(slope of line a).
* the reflection has the opposite of the original slope. i.e., (slope of reflection) = -(slope of line a)
if the original slope of line a is 1 or -1, those two will be the same. (as i advised above, you can also just draw a picture with those slopes, and watch what happens.)
if the original slope is any other value, then they won't.
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what do you mean by "question type" here? like, multiple choice vs. data sufficiency?ngalinh wrote:This way of explanation is most helpful for us. I wish I can see at least one explanation (like this one) of each question type.
Thanks RON.
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