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In the rectangular coordinate system above, the line y = x [#permalink]
 31 Mar 2012, 02:54
Question Stats:
61% (02:02) correct
38% (00:53) wrong based on 10 sessions
In the rectangular coordinate system above, the line y = x is the perpendicular bisector of segment AB (not shown), and the x-axis is the perpendicular bisector of segment BC (not shown). If the coordinates of point A are (1, 4), what are the coordinates of point C? A. (-4, -1) B. (-1, 4) C. (4, -1) D. (1, -4) E. (4, 1) Any idea guys how to solve this mathematically?
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Re: In the rectangular coordinate system above, the line y = x [#permalink]
31 Mar 2012, 04:26
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enigma123 wrote: In the rectangular coordinate system above, the line y = x is the perpendicular bisector of segment AB (not shown), and the x-axis is the perpendicular bisector of segment BC (not shown). If the coordinates of point A are (1, 4), what are the coordinates of point C?
A. (-4, -1)
B. (-1, 4)
C. (4, -1)
D. (1, -4)
E. (4, 1) Since the line y=x is the perpendicular bisector of segment AB, then the point B is the mirror reflection of point A around the line y=x, so its coordinates are (4, 1). The same way, since the x-axis is the perpendicular bisector of segment BC then the point C is the mirror reflection of point B around the x-axis, so its coordinates are (4, -1). Answer: C. The question becomes much easier if you just draw rough sketch of the diagram: Attachment: 
graph.png [ 12.57 KiB | Viewed 2256 times ]
Now, you can simply see that options A, B, and D (blue dots) just can not be the right answers. As for option E: point (4, 1) coincides with point B, so it's also not the correct answer. Only answer choice C remains. Answer: C.
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Re: In the rectangular coordinate system above, the line y = x [#permalink]
01 Aug 2012, 05:04
What is the mirror image of a point (x,y) around Y-axis?
Also what is the mirror image of a point (x,y) around line y=-x?
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Re: In the rectangular coordinate system above, the line y = x [#permalink]
01 Aug 2012, 06:56
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teal wrote: What is the mirror image of a point (x,y) around Y-axis?
Also what is the mirror image of a point (x,y) around line y=-x? The mirror image of (x,y) around the Y-axis is (-x,y).For the second question: Assume we have a point P(a,b) and we want to find its mirror image around the line y = -x. Let's denote the point we seek by Q(A,B). See the attached drawing. The equation of the line passing through P and perpendicular to the line y = -x is y - b = x - a, or y = x + b - a. Since Q is also on this line, we have B = A + b - a, from which A - B = a - b. The middle point of the line segment PQ (denoted by M) is also on the line y = -x, therefore \frac{a+A}{2}=-\frac{b+B}{2}, or A + B = -a - b. Solving for A and B, we find that A = -b, B =-a. Therefore, the mirror image of (x,y) around the line y = -x is (-y, -x).
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Re: In the rectangular coordinate system above, the line y = x [#permalink]
02 Jan 2013, 05:09
enigma123 wrote: In the rectangular coordinate system above, the line y = x is the perpendicular bisector of segment AB (not shown), and the x-axis is the perpendicular bisector of segment BC (not shown). If the coordinates of point A are (1, 4), what are the coordinates of point C?
A. (-4, -1)
B. (-1, 4)
C. (4, -1)
D. (1, -4)
E. (4, 1)
Any idea guys how to solve this mathematically? For me, the best approach to this question is to draw and estimate AB and BC lines. Doing so obviously shows that:
(a) y has negative coordinates... Thus, eliminate B and E.
(b) x has positive coordinates beyond 2. Thus, eliminate A and D.
Answer: C Or, if you want to be really sure... We can get the line perpendicular x=y. (a) get negative reciprocal of slope of y=x which is m=1. Thus, reciprocal is m=-1. Perpendicular line: y=-x + b (b) calculate b using A coordinates: y = -x + b ==> 4 = -1 + b ==> b = 5 (c) get the pt. of intersection. -x + 5 = x ==> x = 2.5 (d) get y=-(2.5) + 5 --> y = 2.5 So, obviously... C would have negative for y coordinate and x > 2.5... only C fits the bill Answer: C But still, drawing should suffice...
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Re: In the rectangular coordinate system above, the line y = x
[#permalink]
02 Jan 2013, 05:09
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