Understand transformations in the plane - HiSET
Card 0 of 164
The graph on the left shows an object in the Cartesian plane. A transformation is performed on it, resulting in the graph on the right.
Which of the following transformations best fits the graphs?
The graph on the left shows an object in the Cartesian plane. A transformation is performed on it, resulting in the graph on the right.
Which of the following transformations best fits the graphs?
A dilation is the stretching or shrinking of a figure.
A rotation is the turning of a a figure about a point.
A reflection is the flipping of a figure across a line.
A translation is is the sliding of a figure in a direction.
With a translation, the image is not only congruent to its original size and shape, but its orientation remains the same. A translation fits this figure best because the shape seems to move upward and rightward without changing size, shape, or orientation.
A dilation is the stretching or shrinking of a figure.
A rotation is the turning of a a figure about a point.
A reflection is the flipping of a figure across a line.
A translation is is the sliding of a figure in a direction.
With a translation, the image is not only congruent to its original size and shape, but its orientation remains the same. A translation fits this figure best because the shape seems to move upward and rightward without changing size, shape, or orientation.
Compare your answer with the correct one above
Consider regular Hexagon 
.
On this hexagon, perform the translation 
. Then perform a 
clockwise rotation on the image with center at 
.
Let 
be the image of 
under these transformations, 
be the image of 
, and so forth. Under these images, which point on the original hexagon does 
fall?
Consider regular Hexagon 
On this hexagon, perform the translation 
Let 
The translation 
on a figure is the translation that shifts a figure so that the image of 
coincides with 
. All other points shift the same distance in the same direction. Below shows the image of the given hexagon under this translation, with the image of 
marked as 
:
If this new hexagon is rotated clockwise 
- one third of a turn - about 
- the image is the original hexagon, but the vertices can be relabeled. Letting 
be the image of 
under this rotation, and so forth:
coincides with 
in the original hexagon, making 
the correct response.
The translation 
If this new hexagon is rotated clockwise 
Compare your answer with the correct one above
Consider regular Hexagon 
.
On this hexagon, perform the translation 
. Then perform a 
rotation on the image with center at 
.
Let 
be the image of 
under these transformations, 
be the image of 
, and so forth. Under these images, which point on the original hexagon does 
fall?
Consider regular Hexagon 
On this hexagon, perform the translation 
Let 
The translation 
on a figure is the translation that shifts a figure so that the image of coincides with 
. All other points shift the same distance in the same direction. Below shows the image of the given hexagon under this translation, with 
the image of 
:
If this new hexagon is rotated 
- one half of a turn - about 
- the image is the original hexagon, but the vertices can be relabeled. Letting 
be the image of 
under this rotation, and so forth:
coincides with 
in the original hexagon, making 
the correct response.
The translation 
If this new hexagon is rotated 
Compare your answer with the correct one above
Consider regular Hexagon 
.
On this hexagon, perform the translation 
. Then perform a 
rotation on the image with center at 
. Let 
be the image of 
under these transformations, and so forth.
Which of the following correctly shows Hexagon 
relative to Hexagon ?
Consider regular Hexagon
On this hexagon, perform the translation 
Which of the following correctly shows Hexagon 
The translation 
on a figure is the translation that shifts a figure so that the image of 
, which we will call 
, coincides with 
. All other points shift the same distance in the same direction. Below shows the image of the given hexagon under this translation:
If this new hexagon is rotated clockwise - one third of a turn - about , and call 
the image of 
, and so forth, the result is as follows:
Removing the intermediate markings, we see that the correct response is
The translation 
If this new hexagon is rotated clockwise 
Removing the intermediate markings, we see that the correct response is
Compare your answer with the correct one above
Consider regular Hexagon .
On this hexagon, perform the translation 
. Then reflect the hexagon about 
. Let 
be the image of 
under these transformations, and so forth.
Which point on Hexagon is the image of 
under these transformations?
Consider regular Hexagon
On this hexagon, perform the translation 
Which point on Hexagon 
The translation 
on a figure is the translation that shifts a figure so that the image of 
, which we will call 
, coincides with 
. All other points shift the same distance in the same direction. Below shows the image of the given hexagon under this translation, with the image of 
marked as 
:
If the image is reflected about 
, the new image is the original hexagon. Calling 
the image of 
under this reflection, we get the following:
, the image of 
under these two transformations, coincides with 
.
The translation 
If the image is reflected about 
Compare your answer with the correct one above
Translate the graph of the equation
left four units and down six units. Give the equation of the image.
Translate the graph of the equation
left four units and down six units. Give the equation of the image.
If the graph of an equation is translated to the right 
units, and upward 
units, the equation of the image can be found by replacing 
with 
and 
with 
in the equation of the original graph.
Since we are moving the graph of the equation
left four units and down six units, we set 
and 
; we can replace 
with 
, or 
, and 
with 
, or 
. The equation of the image can be written as
Simplify by distributing:
Collect like terms:
Add 26 to both sides:
,
the correct choice.
If the graph of an equation is translated to the right 
Since we are moving the graph of the equation
left four units and down six units, we set 
Simplify by distributing:
Collect like terms:
Add 26 to both sides:
the correct choice.
Compare your answer with the correct one above
Translate the graph of the equation
right two units and up five units. Give the equation of the image.
Translate the graph of the equation
right two units and up five units. Give the equation of the image.
If the graph of an equation is translated to the right 
units, and upward 
units, the equation of the image can be found by replacing 
with 
and 
with 
in the equation of the original graph.
Since we are moving the graph of the equation
right two units and up five units, we set 
and 
; we can therefore replace 
with 
and 
with 
. The equation of the image can be written as
This can be rewritten by applying the binomial square pattern as follows:
Collect like terms; the equation becomes
Subtract 100 from both sides:
If the graph of an equation is translated to the right 
Since we are moving the graph of the equation
right two units and up five units, we set 
This can be rewritten by applying the binomial square pattern as follows:
Collect like terms; the equation becomes
Subtract 100 from both sides:
Compare your answer with the correct one above
Translate the graph of the equation
right four units and down two units. Give the equation of the image.
Translate the graph of the equation
right four units and down two units. Give the equation of the image.
If the graph of an equation is translated to the right 
units, and upward 
units, the equation of the image can be found by replacing 
with 
and 
with 
in the equation of the original graph.
Since we are moving the graph of the equation
right four units and down two units, we set 
and 
; we can therefore replace 
with 
, and 
with 
, or 
. The equation of the image can be written as
The expression at right can be simplified. First, use the distributive property on the middle expression:
Now, simplify the first expression by using the binomial square pattern:
Collect like terms on the right:
Subtract 2 from both sides:
,
the equation of the image.
If the graph of an equation is translated to the right 
Since we are moving the graph of the equation
right four units and down two units, we set 
The expression at right can be simplified. First, use the distributive property on the middle expression:
Now, simplify the first expression by using the binomial square pattern:
Collect like terms on the right:
Subtract 2 from both sides:
the equation of the image.
Compare your answer with the correct one above
Translate the graph of the equation
left three units and down five units. Give the equation of the image.
Translate the graph of the equation
left three units and down five units. Give the equation of the image.
If the graph of an equation is translated to the right 
units, and upward 
units, the equation of the image can be found by replacing 
with 
and 
with 
in the equation of the original graph.
Since we are moving the graph of the equation
left three units and down five units, we set 
and 
; we can therefore replace 
with 
, or 
, and 
with 
, or 
. The equation of the image can be written as
We can simplify the expression on the right by distributing:
Collect like terms:
Subtract 5 from both sides:
,
the correct equation of the image.
If the graph of an equation is translated to the right 
Since we are moving the graph of the equation
left three units and down five units, we set 
We can simplify the expression on the right by distributing:
Collect like terms:
Subtract 5 from both sides:
the correct equation of the image.
Compare your answer with the correct one above
On the coordinate plane, let 
, 
, and 
be located at the origin, 
, and 
. Construct the median of 
from 
and let the foot of the median be 
. On the triangle, perform the translation 
. Where is the image of 
?
On the coordinate plane, let 
By definition, a median of a triangle has as its endpoints one vertex and the midpoint of the opposite side. Therefore, the endpoints of the median from 
are 
itself, which is at 
, and 
, which itself is the midpoint of the side with origin 
and 
, which is 
, as its endpoints.
The midpoint of a segment with endpoints at 
and 
is located at
,
so, substituting the coordinates of 
and 
in the formula, we see that 
is
, or 
.
See the figure below:
To perform the translation 
, or, equivalently,
,
on a point, it is necessary to add
and
to the 
- and 
- coordinates, respectively. Therefore, the image of 
is located at
,
or
.
By definition, a median of a triangle has as its endpoints one vertex and the midpoint of the opposite side. Therefore, the endpoints of the median from 
The midpoint of a segment with endpoints at
so, substituting the coordinates of 
See the figure below:
To perform the translation 
on a point, it is necessary to add
and
to the 
or
Compare your answer with the correct one above
On the coordinate plane, let 
and 
be located at 
and 
, respectively. Let 
be the midpoint of 
and let 
be the midpoint of 
. On the segment, perform the translation 
. Where is the image of 
located?
On the coordinate plane, let 
The midpoint of a segment with endpoints at 
and 
is located at
Substitute the coordinates of 
and 
in this formula to find that midpoint 
of 
is located at
, or 
.
Substitute the coordinates of 
and 
to find that midpoint 
of 
is located at
, or 
.
To perform the translation 
, or, equivalently,
,
on a point, it is necessary to add
to its 
-coordinate, and
to its 
-coordinate.
Therefore, the 
-coordinate of the image of 
under this translation is
;
its 
-coordinate is
The image of 
is located at 
.
The midpoint of a segment with endpoints at 
Substitute the coordinates of 
Substitute the coordinates of 
To perform the translation 
on a point, it is necessary to add
to its 
to its 
Therefore, the 
its 
The image of 
Compare your answer with the correct one above
What is the result of reflecting the point 
over the y-axis in the coordinate plane?
What is the result of reflecting the point
Reflecting a point
over the y-axis geometrically is the same as negating the x-coordinate of the ordered pair to obtain
.
Thus, since our initial point was
and we want to reflect it over the y-axis, we obtain the reflection by negating the first term of the ordered pair to get
.
Reflecting a point
over the y-axis geometrically is the same as negating the x-coordinate of the ordered pair to obtain
Thus, since our initial point was
and we want to reflect it over the y-axis, we obtain the reflection by negating the first term of the ordered pair to get
Compare your answer with the correct one above
How many lines of symmetry does the above figure have?
How many lines of symmetry does the above figure have?
A line of symmetry of a figure is about which the reflection of the figure is the figure itself. The diagram below shows the only line of symmetry of the figure.
The correct response is one.
A line of symmetry of a figure is about which the reflection of the figure is the figure itself. The diagram below shows the only line of symmetry of the figure.
The correct response is one.
Compare your answer with the correct one above
On the coordinate plane, the point 
is reflected about the 
-axis. The image is denoted 
. Give the distance 
.
On the coordinate plane, the point 
The reflection of the point at 
about the 
-axis is the point at 
; therefore, the image of is 
. Since these two points have the same 
-coordinate, the distance between them is the absolute value of the difference between their 
-coordinates:
The reflection of the point at 
Compare your answer with the correct one above
Consider regular Hexagon 
; let 
and 
be the midpoints of 
and 
. Reflect the hexagon about 
, then again about 
. With which of the following points does the image of 
under these reflections coincide?
Consider regular Hexagon 
Refer to the figure below, which shows the reflection of 
about 
; we will call this image 
.
Note that 
coincides with 
. Now, refer to the figure below, which shows the reflection of 
about 
; we will call this image - the final image - 
Note that 
coincides with 
, making this the correct response.
Refer to the figure below, which shows the reflection of 
Note that 
Note that 
Compare your answer with the correct one above
Consider regular Hexagon 
; let 
and 
be the midpoints of 
and 
. Reflect the hexagon about 
, then again about 
. Which of the following clockwise rotations about the center would result in each point being its own image under this series of transformations?
Consider regular Hexagon 
Refer to the figure below, which shows the reflection of the given hexagon about 
; we will call 
the image of 
, call 
the image of 
, and so forth.
Now, refer to the figure below, which shows the reflection of the image about 
; we will call 
the image of 
, call 
the image of 
, and so forth.
Note that the vertices coincide with those of the original hexagon, and that the images of the points are in the same clockwise order as the original points. Since 
coincides with 
, 
coincides with 
, and so forth, a clockwise rotation of five-sixth of a complete turn - that is, 
,
is required to make each point its own image under the three transformations.
Refer to the figure below, which shows the reflection of the given hexagon about 
Now, refer to the figure below, which shows the reflection of the image about 
Note that the vertices coincide with those of the original hexagon, and that the images of the points are in the same clockwise order as the original points. Since 
is required to make each point its own image under the three transformations.
Compare your answer with the correct one above
On the above right triangle perform a dilation of scale factor 
with the center of the dilation at the orthocenter of the triangle. Let the images of 
, 
, and 
be 
, 
, and 
, respectively.
Which of the following correctly shows 
relative to 
?
On the above right triangle perform a dilation of scale factor 
Which of the following correctly shows 
The orthocenter of a triangle can be located by finding the intersection of the three altitudes of the triangle - the segments connecting each vertex to its opposite side, perpendicular to the respective side. Since the triangle is right, 
and 
are two of the altitudes, which intersect at 
; the third altitude must also pass through 
, since the three altitudes are concurrent. Therefore, we perform a dilation of the triangle with respect to center 
.
This is done by mapping 
and 
to the midpoints of 
and 
, respectively, and by mapping 
to itself. The triangle is seen below:
This figure is the correct choice.
The orthocenter of a triangle can be located by finding the intersection of the three altitudes of the triangle - the segments connecting each vertex to its opposite side, perpendicular to the respective side. Since the triangle is right, 
This is done by mapping 
This figure is the correct choice.
Compare your answer with the correct one above
On the above right triangle perform a dilation of scale factor 
with the center of the dilation at the circumcenter of the triangle. Let the images of 
, 
, and 
be 
, 
, and 
, respectively.
Which of the following correctly shows 
relative to 
?
On the above right triangle perform a dilation of scale factor 
Which of the following correctly shows 
The circumcenter of a triangle can be located by finding the intersection of the perpendicular bisectors of the three sides of the triangle. The perpendicular bisectors are shown below, with point of intersection 
:
It can be seen that, as is characteristic of a right triangle, this point is the midpoint of the hypotenuse. Construct 
. A dilation of scale factor 
with center 
can be performed by letting 
, 
, and 
be the midpoints of 
, 
, and 
, respectively:
Removing the perpendicular bisectors and 
, we see that the correct choice is the figure
The circumcenter of a triangle can be located by finding the intersection of the perpendicular bisectors of the three sides of the triangle. The perpendicular bisectors are shown below, with point of intersection 
It can be seen that, as is characteristic of a right triangle, this point is the midpoint of the hypotenuse. Construct 
Removing the perpendicular bisectors and 
Compare your answer with the correct one above
Consider Square 
. Perform two dilations successively, each with scale factor 
; the first dilation should have center 
, the second, 
. Call the image of 
under these dilations 
; the image of 
, 
, and so forth.
Which of the following diagrams correctly shows Square 
relative to Square 
?
Consider Square 
Which of the following diagrams correctly shows Square 
To perform a dilation with center 
and scale factor 
, find the midpoints of the segments connecting 
to each point, and connect those points. We can simplify the process by finding the midpoints of 
, 
, and 
, and naming them 
, 
, and 
, respectively; 
, the image of center 
, is just 
itself. The figure is below:
Now, do the same thing to the new square, but with 
as the center. The figure is below:
The final image, relative to the original square, is below:
To perform a dilation with center 
Now, do the same thing to the new square, but with 
The final image, relative to the original square, is below:
Compare your answer with the correct one above
On the above right triangle perform a dilation of scale factor 
with the center of the dilation at the centroid of the triangle. Let the images of 
, 
, and 
be 
, 
, and 
, respectively.
Which of the following correctly shows 
relative to 
?
On the above right triangle perform a dilation of scale factor 
Which of the following correctly shows 
The centroid of a triangle can be located by finding the intersection of the three medians of the triangle - the segments that connect each vertex to the midpoint of its opposite side. The medians are shown below, with point of intersection 
:
A dilation of scale factor 
with center 
can be performed by letting 
, 
, and 
be the midpoints of 
, 
, and 
, respectively:
Removing the medians and 
, we see that the correct choice is the figure
The centroid of a triangle can be located by finding the intersection of the three medians of the triangle - the segments that connect each vertex to the midpoint of its opposite side. The medians are shown below, with point of intersection 
A dilation of scale factor 
Removing the medians and 
Compare your answer with the correct one above