Thread: Triangle reflection maths

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

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).

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 . The x-coordinate will not change from P to . What will change is the y-coordinate. If k>b, then the y-coordinate of will be . That is, the y-coordinate of 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

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

EDIT: Also, regarding the 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.

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 , 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 . 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 is

Now, plug (1,1) into .

We get .

Next, we compute the intersection of the lines 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 will not under go any change.

The y-coordinate of will jump up twice the difference between the y-coordinate of A and the y-coordinate of ; the y-coordinate of is .

So,

-Andy