If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it. - Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

Solved Examples: Geometric Transformations

As applicable, verify your answers with the Geometry Transformations Calculators
Prerequisite Topics:
(1.) Relations and Functions (Graph)
(2.) Trigonometry (Unit Circle)
(3.) Trigonometry (Trigonometric Identities)

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

Solve all questions.
You may solve it geometrically by construction, or algebraically by formulas or both.
The method used here is algebraically (by formulas) mainly. The formulas are stated in the Notes.
For some questions, both methods: geometric (by construction) and algebraic (by formulas) methods are used. For those questions, solving it geometrically is highly recommended due to the time constraint in the test.
Show all work.

(1.) ACT Point A is located at (3, 8) in the standard (x, y) coordinate plane.
What are the coordinates of A', the image of A after it is reflected across the y-axis?

$ A.\;\; (3, -8) \\[3ex] B.\;\; (-3, -8) \\[3ex] C.\;\; (-3, 8) \\[3ex] D.\;\; (8, 3) \\[3ex] E.\;\; (-8, 3) \\[3ex] $

For any point, (x, y); reflection across the y-axis gives (-x, y)
This implies

$ A(3, 8) \rightarrow A'(-3, 8) $

(2.) ACT A triangle, $\triangle$ABC, is reflected across the x-axis to have the image $\triangle$A'B'C' in the standard (x, y) coordinate plane: thus, A reflects to A'.
The coordinates of point A are (c, d).
What are the coordinates of point A' ?

$ F.\;\; (c, -d) \\[3ex] G.\;\; (-c, d) \\[3ex] H.\;\; (-c, -d) \\[3ex] J.\;\; (d, c) \\[3ex] K.\;\; Cannot\;\;be\;\;determined\;\;from\;\;the\;\;given\;\;information \\[3ex] $

For any point, (x, y); reflection across the x-axis gives (x, -y)
This implies

$ A(c, d) \rightarrow A'(c, -d) $

(3.) ACT In the standard (x, y) coordinate plane, point A has coordinates (-7, -5).
Point A is translated 7 units to the left and 5 units down, and that image is labeled A'.
What are the coordinates of A'?

$ F.\;\; (-14, -10) \\[3ex] G.\;\; (-12, -12) \\[3ex] H.\;\; (-7, -10) \\[3ex] J.\;\; (0, 0) \\[3ex] K.\;\; (14, 10) \\[3ex] $

Translated to the left or right is only for the x-coordinate
Translated up or down is only for the y-coordinate

$ \underline{Translated\;\;7\;\;units\;\;to\;\;the\;\;left} \\[3ex] A(-7, -5) \rightarrow A'(-7 - 7, -5) \\[3ex] A(-7, -5) \rightarrow A'(-14, -5) \\[3ex] \underline{Translated\;\;5\;\;units\;\;down} \\[3ex] A'(-14, -5) \rightarrow A''(-14, -5 - 5) \\[3ex] A'(-14, -5) \rightarrow A''(-14, -10) $

(4.) ACT A line segment has endpoints (a, b) and (c, d) in the standard (x, y) coordinate plane, where a, b, c, and d are distinct positive integers.
The segment is reflected across the x-axis.
After this reflection, what are the coordinates of the endpoints of the image?

$ F.\;\; (-a, b)\;\;and\;\;(-c, d) \\[3ex] G.\;\; (a, -b)\;\;and\;\;(c, -d) \\[3ex] H.\;\; (-a, -b)\;\;and\;\;(-c, -d) \\[3ex] J.\;\; (a, b)\;\;and\;\;(c, d) \\[3ex] K.\;\; (a, 0)\;\;and\;\;(c, 0) \\[3ex] $

For any point, (x, y); reflection across the x-axis gives (x, -y)
This implies

$ (a, b) \rightarrow (a, -b) \\[3ex] (c, d) \rightarrow (c, -d) \\[5ex] Option\;G:\;\; (a, -b)\;\;and\;\;(c, -d) $

(5.) GCSE Here is a parallelogram.

The parallelogram is translated 4 squares to the left and 3 squares up.
Draw the translated parallelogram.

translated left: only the x-coordinate is affected
translated up: only the y-coordinate is affected

First: Let us draw a coordinate system to enclose the parallelogram so we can find the coordinates of the four vertices. Mark the four vertices as points: A, B, C, and D.

$ \underline{translated\;\;4\;\;squares\;\;left} \\[3ex] A(0,0) \rightarrow A'(0 - 4, 0) \rightarrow A'(-4, 0) \\[3ex] B(3, 0) \rightarrow B'(3 - 4, 0) \rightarrow B'(-1, 0) \\[3ex] C(1, 2) \rightarrow C'(1 - 4, 2) \rightarrow C'(-3, 2) \\[3ex] D(4, 2) \rightarrow D'(4 - 4, 2) \rightarrow D'(0, 2) \\[5ex] \underline{translated\;\;3\;\;squares\;\;up} \\[3ex] A'(-4, 0) \rightarrow A''(-4, 0 + 3) \rightarrow A''(-4, 3) \\[3ex] B'(-1, 0) \rightarrow B''(-1, 0 + 3) \rightarrow B''(-1, 3) \\[3ex] C'(-3, 2) \rightarrow C''(-3, 2 + 3) \rightarrow C''(-3, 5) \\[3ex] D'(0, 2) \rightarrow D''(0, 2 + 3) \rightarrow D''(0, 5) \\[3ex] $ The translated parallelogram is the one in red color.

(6.) ACT In the standard (x, y) coordinate plane below, $\triangle$ABC will be translated 10 units down and then the resulting image will be reflected over the y-axis.
What will be the coordinates of the final image of A resulting from both transformations?

$ A.\;\; (-5, 9) \\[3ex] B.\;\; (-1, 9) \\[3ex] C.\;\; (1, -9) \\[3ex] D.\;\; (5, -10) \\[3ex] E.\;\; (5, -9) \\[3ex] $

Translated up or down is only for the y-coordinate
For any point, (x, y); reflection across the y-axis gives (-x, y)
We are only concerned with Point A

$ \underline{Translated\;\;10\;\;units\;\;down} \\[3ex] A(-5, 1) \rightarrow A'(-5, 1 - 10) \\[3ex] A(-5, 1) \rightarrow A'(-5, -9) \\[5ex] \underline{Reflected\;\;over\;\;y-axis} \\[3ex] A'(-5, -9) \rightarrow A''(5, -9) $

(7.) ACT The 2 circles graphed in the standard (x, y) coordinate plane below are centered at the origin, O.
In coordinate units, the radius of the smaller circle is 2, and the radius of the larger circle is 4.
Points A(-4, 0), B, and C(4, 0) are on the larger circle.
The measure of ∠BOC is 45°.

(Note: Both axes have the same scale.)
A 3rd circle, not shown, is the image resulting from applying the 1st transformation listed below to the smaller circle and then applying the 2nd transformation listed below to the result of the 1st transformation.
1st: A dilation with center O and scale factor 2
2nd: A translation of 8 coordinate units to the right
The 3rd circle has how many points in common with the larger circle?

$ A.\;\; 0 \\[3ex] B.\;\; 1 \\[3ex] C.\;\; 2 \\[3ex] D.\;\; 4 \\[3ex] E.\;\; Infinitely\;\;many \\[3ex] $

$ Radius\;\;of\;\;smaller\;\;circle = 2 \\[3ex] Radius\;\;OE = (2, 0) \\[3ex] \underline{1st\;\;Transformation:\;\;Dilation\;\;with\;\;scale\;\;factor = 2} \\[3ex] Dilation\;\;applies\;\;to\;\;both\;\;x-coordinate\;\;and\;\;y-coordinate \\[3ex] (2, 0) \rightarrow (2 * 2, 2 * 0) \\[3ex] (2, 0) \rightarrow (4, 0) \\[3ex] \implies \\[3ex] OE = OC \\[3ex] 3rd\;\;circle\;(smaller\;\;circle) = larger\;\;circle \\[3ex] \underline{2nd\;\;Transformation:\;\;Translation\;\;of\;\;8\;\;units\;\;to\;\;the\;\;right} \\[3ex] Translation\;\;to\;\;left\;\;or\;\;right\;\;applies\;\;only\;\;to\;\;x-coordinate \\[3ex] (4, 0) \rightarrow (4 + 8, 0) \\[3ex] (4, 0) \rightarrow (12, 0) \\[3ex] Also: \\[3ex] (-4, 0) \rightarrow (-4 + 8, 0) \\[3ex] (-4, 0) \rightarrow (4, 0) \\[3ex] Endpoints\;\;for\;\;3rd\;\;circle = (4,0) \;\;and\;\; (12, 0) \\[3ex] Endpoints\;\;for\;\;larger\;\;circle = (-4, 0)\;\;and\;\;(4, 0) \\[3ex] left\;\;endpoint\;\;of\;\;3rd\;\;circle = right\;\;endpoint\;\;of\;\;larger\;\;circle \\[3ex] $ So, the two circles have only one common point.

(8.) ACT Triangles $\triangle$ABC and $\triangle$A'B'C' are graphed in the standard (x, y) coordinate plane below.

Triangle $\triangle$A'B'C' is the image of $\triangle$ABC under one of the following transformations.
Which one?
F. 90° clockwise rotation about the origin
G. 90° counterclockwise rotation about the origin
H. Horizontal translation
J. Reflection across the x-axis
K. Reflection across the y-axis

Let us write the coordinates: of the points and their images

$ A(-1, 1) \rightarrow A'(1, 1) \\[3ex] B(-4, 7) \rightarrow B'(4, 7) \\[3ex] C(8, 5) \rightarrow C'(-8, 5) \\[3ex] $ For each point, the x-coordinate changed but the y-coordinate did not change.
This implies a reflection across the y-axis

(9.) curriculum.gov.mt Reflect the triangle below in the line x = 1.

First Method: Geometrically
(1.) Draw the line of reflection...already drawn for you: $x = 1$
(2.) Note the horizontal distance between the first vertex (point) of the triangle to the line of reflection: 2 units
(3.) Ensure the same horizontal distance from the line of reflection to the first vertex of the image of the triangle
(4.) Note the other vertices of the triangle and mark the corresponding vertices in the image of the triangle
(5.) Join the sides of the image of the triangle (reflected triangle)

The reflected triangle is:

Second Method: Algebraically
Reflection across the line: x = k (k is a constant) gives (2k - x, y)
This implies: Reflection across the line: x = 1 (1 is a constant) gives [2(1) - x, y]

$ A(-2, 1) \rightarrow A'[2(1) - (-2), 1] \rightarrow A'(2 + 2, 1) \rightarrow A'(4, 1) \\[3ex] B(-2, 4) \rightarrow B'[2(1) - (-2), 4] \rightarrow B'(2 + 2, 4) \rightarrow B'(4, 4) \\[3ex] C(-1, 1) \rightarrow C'[2(1) - (-1), 1] \rightarrow C'(2 + 1, 1) \rightarrow C'(3, 1) \\[3ex] $ The reflected triangle is:

(10.) ACT A point at (-3, 7) in the standard (x, y) coordinate plane is shifted down 3 units and right 7 units.
What are the coordinates of the new point?

$ F.\;\; (-10, 10) \\[3ex] G.\;\; (0, 0) \\[3ex] H.\;\; (4, 4) \\[3ex] J.\;\; (4, 10) \\[3ex] K.\;\; (10, 10) \\[3ex] $

Let the point be A(-3, 7)
shifted down: only the y-coordinate is affected
shifted right: only the x-coordinate is affected

$ \underline{shifted\;\;down\;\;3\;\;units} \\[3ex] A(-3, 7) \rightarrow A'(-3, 7 - 3) \\[3ex] A(-3, 7) \rightarrow A'(-3, 4) \\[5ex] \underline{shifted\;\;right\;\;7\;\;units} \\[3ex] A'(-3, 4) \rightarrow A''(-3 + 7, 4) \\[3ex] A'(-3, 4) \rightarrow A''(4, 4) $

(11.) ACT The graph shown in the standard (x, y) coordinate plane below is to be rotated in the plane 180° about the origin.

One of the following graphs is the result of this rotation.
Which one is it?

Choose two simple points on the initial graph and read the coordinates of those points

$ \theta = 180^\circ \\[3ex] \sin 180 = 0 \\[3ex] \cos 180 = -1 \\[3ex] \underline{First\;\;Point} \\[3ex] (x, y) = (1, 1) \\[3ex] x' = x\cos\theta - y\sin\theta \\[3ex] x' = 1 * \cos 180 - 0 * \sin 180 \\[3ex] x' = 1(-1) - 0(0) \\[3ex] x' = -1 \\[3ex] y' = y\cos\theta + x\sin\theta \\[3ex] y' = 1 * \cos 180 + 1 * \sin 180 \\[3ex] y' = 1(-1) + 1(0) \\[3ex] y' = -1 \\[3ex] (x', y') = (-1, -1) \\[3ex] \underline{Second\;\;Point} \\[3ex] (x, y) = (4, 0) \\[3ex] x' = x\cos\theta - y\sin\theta \\[3ex] x' = 4 * \cos 180 - 0 * \sin 180 \\[3ex] x' = 4(-1) - 0(0) \\[3ex] x' = -4 \\[3ex] y' = y\cos\theta + x\sin\theta \\[3ex] y' = 0 * \cos 180 + 4 * \sin 180 \\[3ex] y' = 0(-1) + 4(0) \\[3ex] y' = 0 \\[3ex] (x', y') = (-4, 0) \\[3ex] $ Review the options for the graph that has (-1, -1) and (-4, 0)
The only graph that has both coordinates is Option D.

(12.) ACT The figure below shows 3 circles in the standard (x, y) coordinate plane.
Which of the following shows the reflection of the 3 circles across the x-axis?

It is highly recommended to solve this question geometrically.
(1.) Reflection is Flipping
(2.) Reflection across the x-axis means same x-values but different y-values
(3.) So, just flip the three circles across the x-axis
It is going to be the same positive x-axis (same x-values)
However, the 2 circles will be the top and the 3rd bigger circle will be below the 2 circles
Option K. is the correct option

(13.) ACT A series of transformations are applied to the graph in the standard (x, y) coordinate plane below.
The graph below is reflected across the x-axis.
The new graph is reflected across the y-axis.
This new graph is rotated 90° clockwise ($\circlearrowright$) about the origin.

The resulting graph is one of the following graphs.
Which one?

For this question, it is highly recommended we do it geometrically.
We shall solve it using both methods: geometrically and algebraically
(Please do not mind my diagram. It is not drawn to scale.) 😊
However, I recommend you do it geometrically for the ACT test because it is a timed test
You have to solve this question in about a minute...so geometric method is recommended.

First Method: Geometrically

Reflect across the x-axis (Flip)

Reflect across the y-axis (Flip)

Rotate 90° about the origin (Turn)

The answer is Option G.

Second Method: Algebraically
Locate two important points in the original diagram
Write the coordinates of both points (Just imagine if it was on a graph paper)
Take note of these two points because we shall look out for them in the final image.

$ \underline{Initial\;\;Points} \\[3ex] A(-3, 1) \\[3ex] B(-1, 4) \\[3ex] \underline{Reflect\;\;across\;\;the\;\;x-axis} \\[3ex] A(-3, 1) \rightarrow A'(-3, -1) \\[3ex] B(-1, 4) \rightarrow B'(-1, -4) \\[3ex] \underline{Reflect\;\;across\;\;the\;\;y-axis} \\[3ex] A'(-3, -1) \rightarrow A''(3, -1) \\[3ex] B'(-1, -4) \rightarrow B''(1, -4) \\[3ex] \underline{Rotate\;\;90^\circ\;\;clockwise} \\[3ex] \theta = -90^\circ...clockwise\;\;rotation \\[3ex] \sin(-90) = -\sin(90) = -1 ...Odd\;\;Identities \\[3ex] \cos(-90) = \cos(90) = 0 ...Even\;\;Identities \\[3ex] x' = x\cos\theta - y\sin\theta \\[3ex] y' = y\cos\theta + x\sin\theta \\[3ex] \underline{Point\;A''(3, -1)} \\[3ex] x = 3 \\[3ex] y = -1 \\[3ex] x' = 3\cos(-90) - (-1)\sin(-90) \\[3ex] x' = 3(0) + 1(-1) \\[3ex] x' = 0 - 1 \\[3ex] x' = -1 \\[3ex] y' = -1\cos(-90) + 3\sin(-90) \\[3ex] y' = -1(0) + 3(-1) \\[3ex] y' = 0 - 3 \\[3ex] y' = -3 \\[3ex] A''(3, -1) \rightarrow A'''(-1, -3) \\[3ex] \underline{Point\;B''(1, -4)} \\[3ex] x = 1 \\[3ex] y = -4 \\[3ex] x' = 1\cos(-90) - (-4)\sin(-90) \\[3ex] x' = 1(0) + 4(-1) \\[3ex] x' = 0 - 4 \\[3ex] x' = -4 \\[3ex] y' = -4\cos(-90) + 1\sin(-90) \\[3ex] y' = -4(0) + 1(-1) \\[3ex] y' = 0 - 1 \\[3ex] y' = -1 \\[3ex] B''(1, -4) \rightarrow B'''(-4, -1) \\[3ex] $ In the options (images), look for the location of those two points
The points should be transformed as $A'''(-1, -3)$ and $B'''(-4, -1)$

The option that has those points is Option G.

(14.) ACT A point with coordinates (a, b) is plotted in the standard (x, y) coordinate plane as shown below.
The point is then reflected across the y-axis.
Which of the following are the coordinates for the point after the reflection?

$ A.\;\; (-a, b) \\[3ex] B.\;\; (a, -b) \\[3ex] C.\;\; (b, a) \\[3ex] D.\;\; (-b, a) \\[3ex] E.\;\; (b, -a) \\[3ex] $

For any point, (x, y); reflection across the y-axis gives (-x, y)
This implies

$ (a, b) \rightarrow (-a, b) $

(18.) ACT For an assignment on symmetry, Crystal created the pattern of digits shown below.
Her teacher commented on the symmetry when evaluating the assignment.
Which of the following is a true comment about the symmetry of this pattern?

A. The pattern has only a horizontal line of symmetry.
B. The pattern has both a horizontal line and a vertical line of symmetry.
C. The pattern has only a vertical line of symmetry.
D. The pattern has a rotational symmetry of 180°
E. The pattern has a rotational symmetry of 90°

Rotational Symmetry: shape looks the same after some rotations from it's initial position
Rotational Symmetry of 180°: shape looks the same when it is rotated about 180°
Okay, please do not laugh at my diagrams 😊

These implies that the pattern has a rotational symmetry of 180°

Top

(24.) ACT A point at (-2, 8) in the standard (x, y) coordinate plane is shifted right 8 units and down 2 units.
What are the coordinates of the new point?

$ F.\;\; (-10, 10) \\[3ex] G.\;\; (0, 0) \\[3ex] H.\;\; (6, 6) \\[3ex] J.\;\; (6, 10) \\[3ex] K.\;\; (10, 10) \\[3ex] $

Let the point be A(-2, 8)
shifted right: only the x-coordinate is affected
shifted down: only the y-coordinate is affected

$ \underline{shifted\;\;right\;\;8\;\;units} \\[3ex] A(-2, 8) \rightarrow A'(-2 + 8, 8) \\[3ex] A(-2, 8) \rightarrow A'(6, 8) \\[3ex] \underline{shifted\;\;down\;\;2\;\;units} \\[3ex] A'(6, 8) \rightarrow A''(6, 8 - 2) \\[3ex] A'(6, 8) \rightarrow A''(6, 6) $

(27.) What are the coordinates of Y?

$ F.\;\; (a, c) \\[3ex] G.\;\; (b, c) \\[3ex] H.\;\; (c, b) \\[3ex] J.\;\; (a + b, c) \\[3ex] K.\;\; (a + c, b) \\[3ex] $

The horizontal distance from $W(0, 0)$ to $X(a, b)$ is $a - 0 = a$
This should be the same horizontal distance from $Z(c, 0)$ to $Y(?, ?)$
However, to get the x-coordinate of $Y$, we should add that horizontal distance to the x-coordinate of $Z$
This means that the x-coordinate of $Y = a + c$
The y-coordinate of $Y$ is the same as that of the y-coordinate of $X$ because they are on the same horizontal plane
So, the y-coordinate of $Y = b$
This implies that the coordinate of $Y$ is $a + c, b$

(28.) The measure of ∠Y is 50°
What is the measure of the angle between $\overline{WX}$ and the y-axis?

$ A.\;\; 35^\circ \\[3ex] B.\;\; 40^\circ \\[3ex] C.\;\; 45^\circ \\[3ex] D.\;\; 50^\circ \\[3ex] E.\;\; 55^\circ \\[3ex] $

$ \triangle WXK \\[3ex] \angle XWK = \angle W = \angle Y = 50^\circ \\[3ex] ... opposite\;\;\angle s\;\;of\;\;a\;\;parallelogram\;\;are\;\;congruent \\[3ex] \angle XKW = 90^\circ ... y-axis \\[3ex] \angle WXK + \angle XWK + \angle XKW = 180^\circ \\[3ex] ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle WXK + 50 + 90 = 180 \\[3ex] \angle WXK + 140 = 180 \\[3ex] \angle WXK = 180 - 140 \\[3ex] \angle WXK = 40^\circ $

(29.) Parallelogram WXYZ is rotated clockwise ($\circlearrowright$) by 90° about the origin.
At what ordered pair is the image of Z located?

$ F.\;\; (0, -c) \\[3ex] G.\;\; (-c, 0) \\[3ex] H.\;\; (0, 0) \\[3ex] J.\;\; (0, c) \\[3ex] K.\;\; (a, b) \\[3ex] $

First Method: Algebraically

$ \theta = -90^\circ ...rotated\;\;clockwise \\[3ex] \cos\theta = \cos(-90) = \cos(90) = 0 ...Even\;\;Identities \\[3ex] \sin\theta = \sin(-90) = -\sin(90) = -1 ...Odd\;\;Identities \\[5ex] Z(c, 0) \\[3ex] x = c \\[3ex] y = 0 \\[5ex] x' = x\cos\theta - y\sin\theta \\[3ex] x' = c(0) - 0(-1) = 0 + 0 = 0 \\[5ex] y' = y\cos\theta + x\sin\theta \\[3ex] y' = 0(0) + c(-1) = 0 - c = -c \\[5ex] (x', y') = (0, -c) \\[5ex] \underline{Clockwise\;\;Rotation\;\;of\;\;90^\circ} \\[3ex] Z(c, 0) \rightarrow Z'(0, -c) \\[5ex] $ Second Method: Geometrically

$ \underline{Clockwise\;\;Rotation\;\;of\;\;90^\circ} \\[3ex] Z(c, 0) \rightarrow Z'(0, -c) $