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Math Help - Triangle reflection maths

  1. #1
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    Exclamation Triangle reflection maths please help

    Original triangle a- 1,1 b- 2,3 c- 4,3

    A) Reflection in y axis
    B) Reflection in the line y= -1
    C) Reflection y=2x

    Now I think the first one would be A- -1,1 B- -2,3 C- -4,3.
    The second I believe to be A- 1, -3 B-2, -5 C- 4, -5.
    The third can someone please help.
    Can you please tell me if the first two are correct and what the third is. Thanks
    Last edited by harietcress; January 11th 2010 at 11:49 AM.
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  2. #2
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    On a coordinate plane, if a point with coordinates (a,b) is reflected about the y-axis, the reflected point's coordinates are (-a,b).

    Similarly, on a coordinate plane, if a point with coordinates (a,b) is reflected about the x-axis, the reflected point's coordinates are (a,-b).
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  3. #3
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    For the second problem, remember that the line of reflection is half-way between the image and the reflected reflected image.

    More specifically, say you are given a point P=(a,b) on a coordinate plane. When P is reflected about the line y=k, whereby k is a constant, we will call the reflected point  P_r . The x-coordinate will not change from P to  P_r . What will change is the y-coordinate. If k>b, then the y-coordinate of  P_r will be  b+2(k-b) . That is, the y-coordinate of  P_r will be the y-coordinate of P plus twice the distance between y-coordinate of P and k. In similar fashion, if k<b, then the y-coordinate will be shifting downward, i.e., the y-coordinate will be [tex] b-2(b-k) [tex].

    So in your problem, you have a triangle that has points with coordinates A=(1,1), B=(2,3), and C=(4,3). Since -1 is smaller than each of the y-coordinates of the points of the triangle, you will be shifting the y-coordinates of A, B and C downward. I'll do the first. Since point A=(1,1) and -1<1, point  A_r = (1, 1-2[1-[-1]]) = (1, -3)

    Note that I used  A_r simply to denote the point that is the reflection of point A.

    EDIT: Also, regarding the  A_r = (1, 1-2[1-[-1]]) = (1, -3) bit, note that I used brackets in lieu of parenthesis in order to avoid the confusion that could occur when differentiating between coordinates-wrapping parenthesis and multiplication parenthesis.
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  4. #4
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    For your third problem, again, I will get you started and show you how to compute the coordinates of the reflection of the first point, then you do the other two points.

    We have point A=(1,1), and we want to determine the coordinates of  A_r , which is the point you get when you reflect point A over y=2x. We want to find the point that is on the opposite side of the line y=2x AND is equidistant along with point A to y=2x. How do we do this? First, let's compute the slope of the equation of the line perpendicular to  y=2x . The slope of a perpendicular line takes half the negative reciprocal of the slope of the original line. Thus, the slope of the line perpendicular to  y=2x is  y=-\frac{1}{2}.

    Now, plug (1,1) into  y=-\frac{1}{2}x + b .

    We get   1 = b - \frac{1}{2} \implies b=\frac{3}{2} .

    Next, we compute the intersection of the lines  y=2x and [mth] y=-\frac{1}{2}x + \frac{3}{2} [/tex], which is (1,2).

    Since the x-coordinate of A is the same as the x-coordinate of (1,2), the x-coordinate of  A_r will not under go any change.

    The y-coordinate of  A_r will jump up twice the difference between the y-coordinate of A and the y-coordinate of  A_r ; the y-coordinate of  A_r is  1+2(2-1) = 3 .

    So,  A_r = (1,3).

    -Andy
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