If a line is reflected across the line y=x, its slope is the reciprocal of the original slope. So, for y=mx+n to be parallel to y=px+q, it must be true that m=1/p (as long as p is not zero).

Statement 1: m=p+2. It is easy to pick values of p that make m=1/p false. For example, p=2 gives m=4, and 4 is not equal to 1/2. So we have to determine if there are ANY cases where m=p+2 AND m=1/p. Set the equations equal to each other:

p+2=1/p
p^2+2p=1
p^2+2p-1=0

the discriminant of this equation is b^2-4ac=8. That is positive, so the equation does have real solutions. Thus there are values of p such that m=p+2 and m=1/p. INSUFFICIENT.

Statement 2: m=3p. once again, it is easy to find values of p that don't work. p=2 makes m=6, and 6 is not equal to 1/2. Solving m=3p and m=1/p simultaneously gives us

3p=1/p
3p^2=1
p^2=1/3
p=+-√3/3

So, INSUFFICIENT.

Statements 1 & 2 together: 3p=p+2, 2p=2, p=1. m=3. Does m=1/p? 3 does not equal 1, so no. So they are definitely not parallel. SUFFICIENT.

Ans: C

I did not get what you explained...
Can you please tell me what exactly is meant by A line with the equation y = px + q is reflected over the line y = x. ??

Sorry. Look at this graph:

https://gmatmathpro.com/wp-content/uploa ... overyx.jpg

It plots a line, y=3x+2 that is reflected across the line y=x. Notice the symmetry the two lines have about the line y=x. It's like y=x is a mirror and one line is the reflection of the other line in the mirror. Reflecting a point across the line y=x essentially is the same as reversing the x and y coordinates of the point. So, (5,1) reflected across y=x becomes (1,5). (9,-3) becomes (-3,9).

If we want to do that for a whole line, we just switch the x and y variables in the equation. In the example of the graph, we start with y=3x+2.

y=3x+2
x=3y+2 (switch the x and y)
And then solve for y:

3y=x-2
y=x/3-2/3

SO that is how i got the other equation to graph.

If we refect y=px+q across the line y=x:

y=px+q
x=py+q (switch the x and y)
py=x-q
y=x/p-q/p (solve for y)

So the slope of this line is 1/p. For this line to be parallel to y=mx+n, the slopes must be equal, so m=1/p must be true.

Oh ok..Thanks for such a detailed explanation...
Now, I want to know what happens if the line is reflected across the line y = -x?
I guess, the point (-a, b) would become (-b,a)...
Takeaways:
The point or line reflecting across the line y = x would have x and y coordinate reversed...(x, y) will become (y,x). y = mx + c will be come x = my + c
The point reflecting across the line y = -x would have x and y coordinate reversed... (-x, y) will become (-y, x)... Please confirm these points...
What would the equation of the line become if the line reflects across y = -x??

Let me preface this by saying I seriously doubt you'll be tested on this on the GMAT, but...

You've got the right idea. If you reflect a point (a,b) across the line y=-x it becomes (-b,-a). That is, The x-coordinate is replaced by the opposite y-coordinate and the y-coordinate is replaced by the opposite x-coordinate. (2,3) becomes (-3,-2) (-3,5) becomes (-5,3) And, as you correctly state, (-a,b) becomes (-b,a).

So, applying this idea to a line, let's say the equation is y=3x+2. As stated above, the y is replaced with the opposite x, and the x is replaced with the opposite y, so the reflection becomes

-x=3(-y)+2
-x=-3y+2
-3y=-x-2
y=x/3+2/3

If you'd like, go to www.graph.tk or use a graphing calculator and graph the lines

y=3x+2
y=-x
y=x/3+2/3

to see that this is correct.