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In the rectangular coordinate system above, the line y = x [#permalink]

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31 Mar 2012, 02:54

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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)

Re: In the rectangular coordinate system above, the line y = x [#permalink]

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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 15487 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.

Re: In the rectangular coordinate system above, the line y = x [#permalink]

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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 \((-y, -x)\).

Attachments

MirrorImage.jpg [ 18.1 KiB | Viewed 14892 times ]

_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: In the rectangular coordinate system above, the line y = x [#permalink]

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02 Jan 2013, 05:09

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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)

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... _________________

Re: In the rectangular coordinate system above, the line y = x [#permalink]

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28 Jun 2013, 00:31

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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)

Let the co-ordinates of B = (p,q). As x=y is the perpendicular bisector of the line segment AB, thus the middle point of AB will lie on x=y itself. Thus, for A(1,4) and B(p,q)-->

\(\frac {p+1} {2} = \frac {q+4} {2}\) --> p-q = 3

Also, the slope of the line segment would be -1--> \(\frac {q-4}{p-1} = -1\) --> \(p+q = 5\).Thus, on solving, the co-ordinates of B (4,1).

Similarly, as y=0(the x-axis) is the perpendicular bisector of BC, thus, the mid point of BC would like on the x-axis and thus, the y co-ordinate of C has to be -1. Thus, only A and C survive. Again, the slope of line segment BC has to be undefined as it is parallel to the y-axis(or perpendicular to the x axis).

The line Y = X always makes 45 deg angle with the X axis and has slope 1

Since the line Y = X is perpendicular to line AB, The slope of AB must be -1 --------[The product of the slopes of a line and its perpendicular is always -1]

Hereinafter, Even if we are not familiar with 'mirror image' concept we can draft the following figure and can check the answer options.

Since X axis itself is bisector of line BC we can deduce that X value of C can not be negative. Eliminate A, B

We also know Y value of C can not be positive. Eliminate E

Now consider the point A(1,4) This point is on the line that has slope -1, so the X value of its opposite end (i.e. X of B and C also) must be greater than 1. So Choice D (1, -4) Can not be the location of C. Eliminate.

Only option left is C, which is the Answer. _________________

Re: In the rectangular coordinate system above, the line y = x [#permalink]

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06 Aug 2014, 02:37

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In the rectangular coordinate system above, the line y = x [#permalink]

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29 Nov 2014, 21:36

EvaJager wrote:

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 \((-y, -x)\).

Hi I have one question. What if the question ask to find the mirror image of (x,y) around the line y = 2x + 3 for example, how to solve it?

Re: In the rectangular coordinate system above, the line y = x [#permalink]

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12 Jan 2016, 08:19

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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