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

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Question Stats: 71% (01:41) correct 29% (02:18) wrong based on 576 sessions

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Attachment: Capture.GIF [ 3.28 KiB | Viewed 38479 times ]
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)

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

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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).

The question becomes much easier if you just draw rough sketch of the diagram:
Attachment: graph.png [ 12.57 KiB | Viewed 38454 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.

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in-the-rectangular-coordinate-system-the-line-y-x-is-the-132646.html
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the-line-represented-by-the-equation-y-4-2x-is-the-perpendi-87573.html

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

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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 37690 times ]

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

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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).

Only option C survives.

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

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

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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.

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

But still, drawing should suffice...
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Re: In the rectangular coordinate system above, the line y = x  [#permalink]

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enigma123 wrote:
Attachment:
Capture.GIF
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)

This problem does contain a diagram that looks like the following: We are given that the line y = x is a perpendicular bisector of line segment AB. This indicates that point B is a reflection of point A across the line y = x. When we reflect a point (a,b) across the line y = x, the reflected point has the coordinates reversed: (b,a). Thus, since point A is at (2,3), point B must be (3,2).

We are next given that the x-axis is a perpendicular bisector of line segment BC. This means that point C must have the same x-coordinate as point B (3) but the opposite y-coordinate of point B (-2). To further elaborate, we can draw a diagram. _________________

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

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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]

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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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

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Hi Bunuel,
Is it always true that a perpendicular bisector will have mirror reflection of the ends?
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Re: In the rectangular coordinate system above, the line y = x  [#permalink]

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NaeemHasan wrote:
Hi Bunuel,
Is it always true that a perpendicular bisector will have mirror reflection of the ends?

Yes, it is always true.
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Re: In the rectangular coordinate system above, the line y = x  [#permalink]

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I took the approach of eliminating the options.

Since X-axis is perpendicular bisector of line BC, it means that B will line in I quadrant and C in IV quadrant . This means for point C x-cordinate will be +ve and y-co-ordinate will be negative. With this finding , remove options A,B and E

Now as per option D if point C is (1,-4) then point B will be (1,+4). But (1,4) is co-ordinate of point A.

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

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