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

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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?

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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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.

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

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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.