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OG 11-199 > Reflection of a point thru line

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OG 11-199 > Reflection of a point thru line [#permalink]

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New post 16 Jul 2009, 20:58
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Only for those who have OG 11 :) I dont want to create a diagram. This post is to discuss about a general concept.

The explanation says something about "Reflection of point A through line y=x". Since B is the reflection co-ordinates of B can be derived by swapping x and y coordinates of A.

This looks very logical however, I think such reflection rule can be applied only to X-axis, Y-axis or lines that make a 45 degrees angle with the axes ( of course the co-ordinates and signs can change based on the quadrant ).

Is my understanding correct? or is this reflection rule generic for a line with any slope?

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Re: OG 11-199 > Reflection of a point thru line [#permalink]

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New post 18 Jul 2009, 09:42
This reflection rule is generic for a line with any slope.

If you know how to work well with distances in the coordinate plane, you will not have to much trouble.

You must calculate the distance between the given point to the line, and then with that same distance, you will find the reflection point.

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Re: OG 11-199 > Reflection of a point thru line [#permalink]

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New post 18 Jul 2009, 12:43
Economist wrote:
Is my understanding correct? or is this reflection rule generic for a line with any slope?


It's certainly not generic - finding points of reflection is much more difficult if you aren't dealing with a simple line like y = x, y = -x, y = 0 or x = 0. It's not something you'll likely ever need to do on a real GMAT. Just to see how much work is involved, let's take a pretty simple line and point:

What are the co-ordinates of the point obtained when (4,0) is reflected in the line y = 2x?

Of course, depending on the answer choices, all you might need to do here is draw the picture, make a good estimate, and if only one answer is close, you're done. That can be a very good approach on some coordinate geometry questions, and will often at least let you rule out a few wrong answer choices. We can work out the exact co-ordinates:

Draw a line from the point (4,0) through the line y = 2x, which is perpendicular to y = 2x. We want to find the point on this line on the opposite side of y = 2x which is the same distance to y = 2x as (4,0). This new line is perpendicular to y = 2x, so its slope is -1/2. Plugging (4,0) into y = (-1/2)x + b, we find that b = 2. Now, y = (-1/2)x + 2 intersects y = 2x at the point (4/5, 8/5); let's call this point P. To get from (4,0) to (4/5, 8/5), we go up by 8/5, and to the left by 16/5. If we do those steps again but starting from the point (4/5, 8/5), we'll find a point on the line y = (-1/2)x + 2 which is as far from P as (4,0) is. So our reflection point is (4/5 - 16/5, 8/5 + 8/5) = (-12/5, 16/5). Or you could use the distance formula to complete this last step, though that leads to a more complicated calculation.
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Re: OG 11-199 > Reflection of a point thru line [#permalink]

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New post 18 Jul 2009, 13:07
Oh sorry, When I copy you "This reflection rule is generic for a line with any slope.", and did not notice that you were referring to the fact of just swapping x and y coordinates.

I meant that the principle is the same, but yes, as IanStewart said: "It's not something you'll likely ever need to do on a real GMAT.", because of the work to do that.

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Re: OG 11-199 > Reflection of a point thru line [#permalink]

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New post 22 Jul 2009, 00:50
Concur with Ian, Diagramming would be easier, wasteful to spend so much time calculating the coordinates.

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Re: OG 11-199 > Reflection of a point thru line [#permalink]

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New post 23 Jul 2009, 05:02
IanStewart wrote:
Economist wrote:
Is my understanding correct? or is this reflection rule generic for a line with any slope?


It's certainly not generic - finding points of reflection is much more difficult if you aren't dealing with a simple line like y = x, y = -x, y = 0 or x = 0. It's not something you'll likely ever need to do on a real GMAT. Just to see how much work is involved, let's take a pretty simple line and point:

What are the co-ordinates of the point obtained when (4,0) is reflected in the line y = 2x?

Of course, depending on the answer choices, all you might need to do here is draw the picture, make a good estimate, and if only one answer is close, you're done. That can be a very good approach on some coordinate geometry questions, and will often at least let you rule out a few wrong answer choices. We can work out the exact co-ordinates:

Draw a line from the point (4,0) through the line y = 2x, which is perpendicular to y = 2x. We want to find the point on this line on the opposite side of y = 2x which is the same distance to y = 2x as (4,0). This new line is perpendicular to y = 2x, so its slope is -1/2. Plugging (4,0) into y = (-1/2)x + b, we find that b = 2. Now, y = (-1/2)x + 2 intersects y = 2x at the point (4/5, 8/5); let's call this point P. To get from (4,0) to (4/5, 8/5), we go up by 8/5, and to the left by 16/5. If we do those steps again but starting from the point (4/5, 8/5), we'll find a point on the line y = (-1/2)x + 2 which is as far from P as (4,0) is. So our reflection point is (4/5 - 16/5, 8/5 + 8/5) = (-12/5, 16/5). Or you could use the distance formula to complete this last step, though that leads to a more complicated calculation.

Great Post. Thanks.

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Re: OG 11-199 > Reflection of a point thru line   [#permalink] 23 Jul 2009, 05:02
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