knight247 wrote:Just a clarification that I need. For line y=px+q the slope is p and its reflection will be in the same plane so even the reflection will have a slope of p rite?

If we take the line y = px+q and reflect it over the line y=x, the reflected line will have slope 1/p

Having said that, I'm pretty sure that this question is out-of-scope. I've never seen an official GMAT question that requires one to know that the reflected line (across the line y=x) will have slope 1/p, and I've never seen this concept mentioned in the GMAT curriculum.

Aside: I'm sure that there are people who remember this concept from high school, and for those people the question will not be a big deal. But, if we're interested in answering the question "Is the concept of reflecting lines across the line y=x within the scope of the GMAT?", my answer is no.

Cheers,
Brent

I should add that I think it is within the scope of the GMAT to ask questions where we reflect individual points across the line y=x. (I just think that reflecting lines is out-of-scope)

In general, if the point (x,y) is reflected across the line y=x, the reflected point is at (y,x)
So, for example, if the point (3,1) is reflected across the line y=x, the reflected point is at (3,1).

Cheers,
Brent

knight247 wrote:Hey Brent,
Appreciate ur response. Is there any way that you could help prove that the slope of line y=mx+b reflected over y=x is 1/m? I'm having a hard time understanding how that is so. Or maybe if you could show me any links where I could learn that. Also, Are there any other similar properties or corollaries related to reflections over y=x that are important to know. I mean, besides the two mentioned here. Thanks

Aside: This is pretty esoteric stuff. You certainly don't need to know this rule or this proof for the GMAT.

That said, here's a pretty rudimentary proof:

If the line y=mx+b is reflected across the line y=x, the equation of the reflected line is x=my+b (to find the equation of any reflected line across y=x, just switch the x's and y's)

Now let's find 2 points on the reflected line x=my+b

When y=0, we get x=m(0)+b, which equals b.
So, one point on the line is (b,0)

When y=1, we get x=m(1)+b, which equals m+b.
So, a second point on the line is (m+b,1)

When we use the slope formula to find the slope between these two points, we get:
Slope = [1-0]/[(m+b)-(b)] = 1/m

Cheers,
Brent

knight247 wrote: Are there any other similar properties or corollaries related to reflections over y=x that are important to know. I mean, besides the two mentioned here. Thanks

Sorry, I forgot to address this question.

For the GMAT, all you really need to know about reflections across the line y=x is that when the point (x,y) is reflected across the line y=x, the reflected point is at (y,x)

Cheers,
Brent