Name:__________________________
Learning Packet: E/C
Date:_____ Period:____
Ms. Anderle
Review of Transformations
There will be an exam when we return on transformations (sorry ). I will be
available for extra help in the morning – at 7 so even students who have H period
can come. Relax. It is not difficult, as long as your memorize those formulas! The
exam has 10 multiple choice questions, 2 questions where you have to draw in the
axis of symmetry, 9 transformation questions, and 2 Part II transformation
questions (graphing). If there are any questions, I will be checking my e-mail all
week. So please do not hesitate to e-mail me.
Transformations Definitions & Formula Sheet
Translation:
Slides a figure a certain distance. The shape does not change size.
Mapping Rule: Ta, b(x, y) = (x + a, y + b).
Dilation:
Enlarges or shrinks an image.
Mapping Rule: Dk(x,y) = (kx,ky)
Line Reflection:
An image is “flipped” over a line.
Mapping Rules:
-Reflection over the x-axis
rx-axis(x, y) = (x, -y)
-Reflection over the y-axis
ry-axis(x, y) = (-x, y)
-Reflection over the y=x line
ry = x(x, y) = (y, x)
-Reflection over the y=-x line
ry=-x(x, y) = (-y, -x)
-Reflection over the line y = k (where k is a NUMBER)
ry = k(x, y) = (x, 2k – y)
-Reflection over the line x = k (where k is a NUMBER)
rx = k(x, y) = (2k – x, y)
Point Reflection:
An image is “flipped” over a point.
-Point Reflection in the Origin
rorigin(x, y) = (-x, -y) or rO(x, y) = (-x, -y
-Point Reflection over a given point (h,k) (where h and k are constants)
r(h, k)(x, y) = (2h – x, 2k – y)
Rotation:
An image is “turned” over a point (usually the origin). Rotations are
usually counterclockwise unless otherwise noted. If the image is rotated
CLOCKWISE the degree value becomes negative.
Rotation of 90° and -270°
R90° = R-270°(x,y) = (-y,x)
Rotation of 180° and -180°
R180° = R-180˚(x,y) = (-x,-y)
Rotation of 270 or -90
R270˚ = R-90˚(x,y) = (y,-x)
Glide Reflection:
A glide reflection occurs when a transformation (a slide) and a reflection (a
flip) are performed after each other. It does not matter if the transformation or
the reflection occurs first. A glide reflection is commutative.
Properties of Transformations:
All transformations except for reflection preserve collinearity. Collinearity
is when the letter order of the vertices are in the same order. In a reflection the
order changes from clockwise to counterclockwise. All transformations except for
a dilation preserve distance. Isometry is when a transformation preserves
distance. Therefore, all transformations except for a dilation are isometry!
Part II Questions:
1. The coordinate of the vertices of ΔGEO are G(-2,-4), E(-3,0), and O(-5,2).
Graph and label ΔGEO.
a. Graph and state the coordinates of ΔG’E’O’, the image of ΔGEO after
the transformation ry-axis.
b. Graph and state the coordinates of ΔG”E”O”, the image of ΔG’E’O’
after the transformation ry=-x.
c. Graph and state the coordinates of ΔG”’E”’O”’, the image of ΔG”E”O”
after the transformation D-2
2. ΔDFG has vertices D(2,8), F(1,2), and G(4,6). Graph ΔDFG and its image after
ry=x ◦ T3,3. Is this a glide reflection?
Name:_________________
Learning Packet: E/C
Date:_____ Period:____
Ms. Anderle
Please complete the below packet. We will be reviewing this topic AFTER the exam.
However, you must hand this in to me on Monday. Again, if you have any questions,
please do not hesitate to e-mail me. In addition, this website will also help you:
http://www.regentsprep.org/Regents/math/geometry/GL1/indexGL1.htm.
Locus:
A locus is a set of points that satisfy a given condition. We want to think of a locus
as a “bunch” of points that all do the same thing. We always draw the locus as a
dotted line. There are 5 basic rules for locus.
1. Locus a fixed distance from a point:
The locus a fixed distance from a point is a circle.
Example:
To sketch a locus simply follow the diagram above. Label d the distance and P
whatever point they give you.
When we draw the locus a fixed distance from a point on the coordinate plane (a
graph) we must use the equation of a circle to “name” the locus. The equation of a
circle is: (x – h)2 + (y – k)2 = r2, where (h,k) is the center of the circle – the point,
and r is the radius of the circle – the fixed distance.
Example:
1. Write the equation of the locus 3 units away from point (2,-1).
Equation: (x – 2)2 + (y + 1)2 = 9
2. Write the equation of the locus 2 units away from the origin.
Equation: (x – 0)2 + (y – 0)2 = 4 OR x2 + y2 = 4
***We will draw the graphs in class***
Now Try:
1. Write the equation of locus 4 units away from the point (8,1)
2. Write the equation of the locus 5 units away from the point (-2,1)
3. Write the equation of the locus 3 units away from the point (-1,-1).
4. Sketch the locus 3 inches away from point P.
5. Sketch the locus 7 feet away from point R.
2. Locus Equidistant from Two Points:
The locus equidistant from two points is the perpendicular bisector of the
segment joining those two points together.
Example:
To sketch the locus equidistant from two points, follow the above diagram. In
addition, show how far the line should be drawn. For example, if points P and Q are
10 inches apart, the perpendicular bisector should be 5 inches away from point P
and 5 inches from point Q (in the middle of those points). Draw in a square to
indicate that the locus is perpendicular to the segment.
When we are writing the equation of the locus equidistant from two points – we
must write the equation of the line perpendicular to the segment connecting those
two points. Therefore, we must find the slope of the line segment. Slope = Δy/Δx.
Once we find the slope, to find the slope the slope of the line perpendicular to it,
we get the negative reciprocal (flip the fraction, then negate).
After, we find the slope of the line perpendicular, we then must find the midpoint
of the segment. Remember the midpoint formula!
Then, we use the midpoint and the slope and plug it into this formula:
(y – y1) = m(x – x1)
Where m = slope of the line perpendicular and (x1,y1) = midpoint
Example:
Find the equation of the locus of points equidistant from (0,1) and (4,3).
Step 1: Find the slope of line segment connecting (0,1) and (4,3)
m = (3 – 1)/(4 - 0)
m = 2/4
m=½
Step 2: Find the slope of the line perpendicular to the line segment connecting (0,1)
and (4,3).
The slope of the line perpendicular to a segment is the negative reciprocal of the
slope.
Therefore, the slope is -2/1 or -2.
Step 3: Find the midpoint of the line segment connecting (0,1) and (4,3).
M = (x1 + x2 , y1 + y2)
2
2
M = (0 + 4 , 1 + 3)
2
2
M = (2,2)
Step 4: Write the equation of the locus equidistant from two points
(y – y1) = m(x – x1)
(y – 2) = -2(x – 2)
You can leave the equation like this!
Now Try:
1. Write the equation of the locus equidistant from points (2,1) and (6,1).
2. Write the equation of the locus equidistant from points (-4,5) and (-2,1).
3. Write the equation of the locus equidistant from points (3,1) and (-5,5).
3. Locus: A fixed distance from a line
The locus a fixed distance from a line is the set of two parallel lines.
Example:
To sketch the locus a fixed distance from a line, replace d with how far the locus is
away from the line.
Writing the equation of the locus a fixed distance from a line:
Examples:
1. Write the equation of the locus 4 units away from the line x=2
Step 1: Figure out if the lines should be y= or x=
The lines should be written as x = because in order for them to parallel to
x=2, the lines must also be x=
Step 2: Add the distance to the original line
x=2 + 4
x=6  this is the equation of the first line
Step 3: Subtract the distance from the original line
x=2 – 4
x=-2  this is the equation of the second line
Step 4: Write out a sentence
x=6 and x=-2 is the equation of the locus 4 units away from the line x=2
2. Write the equation of the locus 2 units away from the x-axis.
Step 1: Figure out if the lines should be y= or x=
The lines should be written as y=, since the x-axis is also known as y=0.
Therefore, in order for the locus to be parallel to the x-axis, the lines must
be y=
Step 2: Add the distance to the original line
y=0+2
y = 2  this is the equation of the first line
Step 3: Subtract the distance from the original line
y=0 – 2
y = -2  this is the equation of the second line
Step 4: Write out a sentence
y = 0 and y = -2 is the equation of the locus 2 units away from the x-axis.
Now Try:
1. What is the equation of the locus 5 units away from the line y = -2?
2. What is the equation of the locus 3 units away from the line y = -4?
3. What is the equation of the locus 6 units away from the line x = 1?
4. Locus: Equidistant from Two Parallel Lines
The locus equidistant from two parallel lines is the locus midway between them.
Example:
Writing the Equation of the Locus Equidistant from Two Parallel Lines:
Examples
1. Write the equation of the locus equidistant from the lines x=2 and x=4.
Step 1: Figure out if the locus should be x= or y=
The locus should be x= because the two lines are x=
Step 2: Find the average of the two lines.
(2+4)/2
6/2
3
Step 3: Put steps 1 and 2 together
x=3 is the equation of the locus equidistant from x=2 and x=4
Now Try:
1. Write the equation of the locus equidistant from y=-2 and y=4
2. Write the equation of the locus equidistant from x=5 and x=-3
3. Write the equation of the locus equidistant from y=2 and y=8.
5. Locus: Equidistant from Two Intersecting Lines
The locus equidistant from two intersecting lines is the bisectors of each pair of
vertical angles formed by the lines.
Example:
We will only be sketching the locus of two intersecting lines. To sketch the locus,
simply follow the above picture.
Now Try:
1. Sketch the locus of the intersecting lines below:
If you have any questions on the packet, please email me. janderle@sfponline.org
I hope that everyone stays healthy! 
-Ms. Anderle