Which of the following points is reflect to y=-x at (-2,1)?

==> You can figure out a point reflecting to y=-x by substituting –y to x-coordinate and –x to y-coordinate. Then, (-2,1) --> (-1, -(-2))=(-1,2) is derived.

Hence, the answer is A.
Answer: A

Which of the following points is reflect to y=-x at (-2,1)?

A. (-1, 2) B. (1,-2) C. (2,1) D. (2,-1) E. (1,2)

Okay so I think "C" is the trap answer in this question. (2,1) reflects (-2,1) but we're dealing with y= "-x"

When reflected about line y=-x the x co-ordinate -2 becomes y co-ordinate and y cordinate 1 becomes x co-ordinate.
Since (-2,1) is in 2nd quadrant we interchange signs annd get (-1,2)

So ans: A

Hi Sir ,

Can you give theory regarding these kinds of problems .

Regards,
Arvind

No specific theory on this but you can check similar questions to practice:
https://gmatclub.com/forum/in-the-xy-coo ... 43502.html
https://gmatclub.com/forum/in-the-rectan ... 32646.html
https://gmatclub.com/forum/the-coordinat ... 27769.html
https://gmatclub.com/forum/the-line-repr ... 27770.html
https://gmatclub.com/forum/if-point-a-co ... 41972.html
https://gmatclub.com/forum/in-the-rectan ... 88473.html
https://gmatclub.com/forum/in-the-rectan ... 44774.html
https://gmatclub.com/forum/the-line-repr ... 87573.html
https://gmatclub.com/forum/in-the-recta ... 29932.html

Hope it helps.

Quick Theory: Let the reflected point be (x,y). The line joining points (-2,1) and (x,y) will be perpendicular to the mirror line.

Mirror line => y = -x has slope = -1. Therefore the line joining reflected and actual points will have slope = 1.

Using the slope rise/run form for the line joining original and reflected point we get:

1 = (y - 1)/(x - (-2))

1 = (y-1)/(x+2)

y-1 = x + 2

y = x + 3

Only Option A satisfies this condition

A.

Consider a circle with the centre in origin and radius-end point in (-2,1).
Graph y = -x is a diameter to this circle, so our answer comes from symmetry with respect to this diameter (y = -x).
+you may play with triangular equalities/symmetries and so on:)

your method is brilliant, but since gmat does not allow test takers to bring rulers, ones will find difficult to estimate the quadrant of a point given annoying locations.
Also, your method is correct, but your calculation is incorrect. OA is A.

ScottTargetTestPrep
hello, does your method apply to the general form of any linear equatation y = mx + n ?

Hi chesstitans,

No, it doesn't apply to any line in the form of y = mx + b (for any value of m and b). The formula works here for y = -x because y = -x is a special line (it makes a 45-degree angle with the negative x-axis). The same goes for lines y = x, y = 0, and x = 0 because they are special lines. However, it doesn't apply to any line in general. That said, there is a formula for reflecting a point of a line in the form of y = mx + b. However, that formula is not easy to memorize, so when a problem arises--for example, reflect (1, 2) over y = 3x + 4--we deal with it individually using concepts of coordinate geometry rather than a formula that few people can memorize.

https://magoosh.com/math/coordinate-geo ... ate-plane/

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