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Assignment 13- Applications
Reflecting an image over three parallel lines
Line A, B and C are all parallel to each other. The distance between line A and Line B is equal to the
distance between line B and line C.
The blue shape is reflected across line A to get the pink shape. The pink shape is then reflected across
line B to get the green shape. Finally, the green shape is reflected across line C to get the purple shape.
A simpler wat to describe this relationship of composition of reflections is to get one reflection.
If we take the blue shape and reflect it across line B, we get the purple shape, and if we take the pink
shape and reflect it across line B, we get the green shape. In this composition of reflections with parallel
lines, the middle line can be used to get the final position of the shape.
Reflecting an image over three concurrent lines
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Line A, B and C all meet in the center point. The Blue shape is the original shape. It is then reflected over
line A to get the Pink shape. The pink shape is then reflected across line B to get the green shape. Finally,
the green shape is reflected across line C to get the Purple shape; the final position of the composition
of reflecting across three concurrent lines.
In the image below, I added a circle with the center as the point where all the lines meet. All the
corresponding points of the original shape also lie on the circle. This shows us that to simplify the
composition of three reflections over three concurrent lines we can use a rotation. Using Geogebra, we
was can calculate the angle of rotation, which varies depending on the lines.
Two parallel lines are parallel if and only if their slopes are equal.
→ Two lines are parallel if their slopes are equal.
Parallel lines are just a reflection of each other. By the properties of isometries, the distance is
preserved between the first line and the reflecting line and between the reflecting line and the
reflected parallel line. As well as the one-to-one correspondence and collinearity.
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𝑏−𝑦
𝑓(𝑏)−𝑓(𝑦)
The slope of line 1 is m1=𝑎−𝑥 and the slope of line 2 is m2=𝑓(𝑎)−𝑓(𝑥)
But since the distance is preserved, the distance between (x,y) and (a,b) is the same as the distance
between f(x,y) and f(a,b), then b-y = f(b)-f(y) and a-x=f(a)-f(x). So the slopes are equal.
←If the slopes of two lines are equal then the lines are parallel.
Let Line 1 y=mx+b1 and line 2 y=mx+b2 where m is the same for the two lines and b1≠ b2.
Assume the lines are not parallel
mx+b1=mx+b2
b1= b2 but b1≠ b2. So that must mean that the lines are parallel.
Therefore, two lines are parallel if and only if their slopes are equal.
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Prove that perpendicular lines have opposite reciprocal slopes.
If you start with one line and rotate it 90° it gives you a second line that is perpendicular to the other.
The slope of a line can be calculated as m=
𝑅𝑖𝑠𝑒
.
𝑅𝑢𝑛
4
1
In the first line we have m1= , which is shown by the
𝑅𝑢𝑛
1
triangle. The rotation makes the new triangle’s m to be m= 𝑅𝑖𝑠𝑒 which makes m2= 4 . But in the second
triangle “Rise” is no longer up but going down so that’s where the negative comes from so the new
1
slope is actually m2= − 4. Thus perpendicular lines have opposite reciprocal slopes.