Understand transformations in the plane - HiSET

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Question

Consider Square . Perform two dilations successively, each with scale factor $\frac{1}{2}$ ; 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 ?

Answer

To perform a dilation with center and scale factor $\frac{1}{2}$ , find the midpoints of the segments connecting to each point, and connect those points. We can simplify the process by finding the midpoints of $\overline{AB}$ , $\overline{AC}$ , and $\overline{AD}$ , 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:

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Question

On the above right triangle perform a dilation of scale factor $\frac{1}{2}$ 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 $\bigtriangleup A'B'C'$ relative to $\bigtriangleup ABC$ ?

Answer

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, $\overline{AB }$ and $\overline{CB}$ 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 $\overline{AB }$ and $\overline{CB}$ , respectively, and by mapping to itself. The triangle is seen below:

This figure is the correct choice.

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Question

On the above right triangle perform a dilation of scale factor $\frac{1}{2}$ 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 $\bigtriangleup A'B'C'$ relative to $\bigtriangleup ABC$ ?

Answer

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 $\overline{BO}$ . A dilation of scale factor $\frac{1}{2}$ with center can be performed by letting , , and be the midpoints of $\overline{AO}$ , $\overline{BO}$ , and $\overline{CO}$ , respectively:

Removing the perpendicular bisectors and , we see that the correct choice is the figure

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Question

On the above right triangle perform a dilation of scale factor $\frac{1}{2}$ 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 $\bigtriangleup A'B'C'$ relative to $\bigtriangleup ABC$ ?

Answer

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 $\frac{1}{2}$ with center can be performed by letting , , and be the midpoints of $\overline{AO}$ , $\overline{BO}$ , and $\overline{CO}$ , respectively:

Removing the medians and , we see that the correct choice is the figure

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Question

Consider Square . Perform two dilations successively, each with scale factor $\frac{1}{2}$ ; 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 ?

Answer

To perform a dilation with center and scale factor $\frac{1}{2}$ , find the midpoints of the segments connecting to each point, and connect those points. We can simplify the process by finding the midpoints of $\overline{CA}$ , $\overline{CB}$ , and $\overline{CD}$ , 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 image, relative to the original square, is below:

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Question

On the above obtuse triangle perform a dilation of scale factor $\frac{1}{2}$ 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 $\bigtriangleup A'B'C'$ relative to $\bigtriangleup ABC$ ?

Answer

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 :

Construct $\overline{AO}$ , $\overline{BO}$ , and $\overline{CO}$ . A dilation of scale factor $\frac{1}{2}$ with center can be performed by letting , , and be the midpoints of $\overline{AO}$ , $\overline{BO}$ , and $\overline{CO}$ , respectively:

Removing the perpendicular bisectors and , we see that the correct choice is the figure

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Question

On the above obtuse triangle perform a dilation of scale factor $\frac{1}{2}$ 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 $\bigtriangleup A'B'C'$ relative to $\bigtriangleup ABC$ ?

Answer

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 the line containing its opposite side, perpendicular to the respective side. Extend $\overline{AB}$ and $\overline{AC}$ to $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC}$ ; the altitudes are shown below, with point of intersection :

A dilation of scale factor $\frac{1}{2}$ with center can be performed by letting , , and be the midpoints of $\overline{AO}$ , $\overline{BO}$ , and $\overline{CO}$ , respectively:

Removing the altitudes and , we see that the correct choice is the figure:

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Question

On the above obtuse triangle, perform a dilation of scale factor $\frac{1}{2}$ 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 $\bigtriangleup A'B'C'$ relative to $\bigtriangleup ABC$ ?

Answer

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 $\frac{1}{2}$ with center can be performed by letting , , and be the midpoints of $\overline{AO}$ , $\overline{BO}$ , and $\overline{CO}$ , respectively:

Removing the medians and , we see that the correct choice is the figure

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Question

On the graph of the equation

$\frac{x^{2}}{81}+\frac{y^{2}}{36}= 1$ ,

perform a dilation with center at the origin and with scale factor $\frac{1}{3}$ .

Give the equation of the resulting figure.

Answer

The graph of the equation

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1$

is an ellipse whose center is at the origin; it has a horizontal axis with endpoints at $(\pm a,0)$ and a vertical axis with endpoints at $(0, \pm b)$ . Thus,

$\frac{x^{2}}{81}+\frac{y^{2}}{36}= 1$ ,

or

$\frac{x^{2}}{9^{2}}+\frac{y^{2}}{6^{2}}= 1$

has its center at the origin, and its endpoints at $(\pm 9,0)$ and $(0, \pm 6)$ .

The center of dilation is the center of the ellipse, so the center of the image will also be at the origin. The endpoints of the ellipse will be the points $\frac{1}{3}$ times the distance away from the origin, in the same directions. Since the center of the dilation is the origin, these points can be found by simply multiplying the coordinates by $\frac{1}{3}$ ; the endpoints of the horizontal axis will be

$\left ( \pm 9 \times \frac{1}{3},0 \right )$ ,

or

$\left ( \pm 3,0 \right )$ ,

and the endpoints of the vertical axis will be

$\left ( 0, \pm 6 \times \frac{1}{3} \right )$

or

$(0, \pm 2)$ .

This dilation is shown in the figure below:

Therefore, and , making the equation of the image

$\frac{x^{2}}{3^{2}}+\frac{y^{2}}{2^{2}}= 1$ ,

or

$\frac{x^{2}}{9}+\frac{y^{2}}{4}= 1$ .

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Question

On the graph of the equation

$\frac{x^{2}}{16}+\frac{y^{2}}{9}= 1$ ,

perform a dilation with center at the origin and with scale factor 3.

Give the equation of the resulting image.

Answer

The graph of the equation

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1$

is an ellipse whose center is at the origin; it has a horizontal axis with endpoints at $(\pm a,0)$ and a vertical axis with endpoints at $(0, \pm b)$ . Thus,

$\frac{x^{2}}{16}+\frac{y^{2}}{9}= 1$ ,

or

$\frac{x^{2}}{4^{2}}+\frac{y^{2}}{3^{2}}= 1$ ,

has its center at the origin, and its endpoints at $(\pm 4,0)$ and $(0, \pm 3)$ .

The center of dilation is the center of the ellipse, so the center of the image will also be at the origin. The endpoints of the ellipse will be the points 3 times the distance away from the origin, in the same directions. Since the center of the dilation is the origin, these points can be found by simply multiplying the coordinates by 3; the endpoints of the horizontal axis will be

$(\pm 4 \times 3,0)$ ,

or

$(\pm 12,0)$ ,

and the endpoints of the vertical axis will be

$(0, \pm 3 \times 3)$

or

$(0, \pm 9)$ .

The figure is shown below:

Therefore, and , making the equation of the image

$\frac{x^{2}}{12^{2}}+\frac{y^{2}}{9^{2}}= 1$ ,

or

$\frac{x^{2}}{144}+\frac{y^{2}}{81}= 1$ .

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Question

On the graph of the equation

$x^{2}+ y^{2} = 64$ ,

perform a dilation with center and scale factor $\frac{1}{3}$ .

Give the equation of the resulting circle.

Answer

The graph of the equation

$(x-h)^{2}+(y-k)^{2} = r^{2}$

is a circle with center and radius .

$x^{2}+ y^{2} = 64$ ,

or

$(x-0) ^{2}+ (y-0)^{2} = 8^{2}$

is a circle with center at origin and radius 8.

A dilation of a circle with scale factor $\frac{1}{3}$ will result in multiplying that radius by $\frac{1}{3}$ , so the radius of the circle will be $8\cdot \frac{1}{3}= \frac{8}{3}$ .

To find the center of the image, note that the origin is 4 units below . The center of the new circle must be $4 \cdot \frac{1}{3} = \frac{4}{3}$ units below , so this center will be $\left ( 0,4- \frac{4}{3} \right )$ , or $\left ( 0, \frac{8}{3} \right )$ . See the figure below:

Substituting in the circle formula, this is

$(x-0)^{2}+\left (y- \frac{8}{3} \right )^{2} = \left (\frac{8}{3} \right )^{2}$ ,

or

$x^{2}+\left (y- \frac{8}{3} \right )^{2} = \frac{64}{9}$ .

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Question

On the coordinate plane, Quadrilateral has its vertices at the following four points:

$A \left(-7,\ -6\right),B\left(-2,8\right),C \left(4,\ 7\right),D\left(8,-4\right)$

Perform a dilation of this quadrilateral with center at the origin and scale factor $\frac{2}{5}$ . Call the image Quadrilateral , where is the image of , etc.

Which of the following choices does not match the point with its correct coordinates?

Answer

The image of a point under a dilation with center and scale factor is the point in the same direction from as , but $s \cdot OP$ units away, where is the distance from to . Since the center of the dilation is the origin , the coordinates of the image of , which is , can be found by multiplying scale factor $\frac{2}{5}$ by each of those of :

$A \left(-7,\ -6\right)$

$\left(-7 \cdot \frac{2}{5},\ -6 \cdot \frac{2}{5}\right)$

$\left(-\frac{14}{5},\ -\frac{12}{5}\right)$

$\left(-2\frac{4}{5},\ -2\frac{2}{5}\right)$

The images of the other three vertices can be found similarly:

$B\left(-2,8\right)$

$\left(-2\cdot \frac{2}{5},8\cdot \frac{2}{5} \right)$

$\left(- \frac{4}{5},- \frac{16}{5} \right)$

$C \left(4,\ 7\right)$

$\left(4\cdot \frac{2}{5} ,\ 7\cdot \frac{2}{5} \right)$

$\left( \frac{8}{5} , \frac{14}{5} \right)$

$\left(\textbf{1} \frac{\textbf{3}}{\textbf{5}} , 2\frac{4}{5} \right)$

$D\left(8,-4\right)$

$\left(8\cdot \frac{2}{5},-4\cdot \frac{2}{5}\right)$

$\left(3 \frac{1}{5},-1 \frac{3}{5}\right)$

Note that only in the case of , the ordered pair in the choice differs (in abscissa) from the correct ordered pair. This makes the correct choice $C \left(1\frac{2}{5},\ 2\frac{4}{5}\right)$ .

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Question

What is the result of reflecting the point over the y-axis in the coordinate plane?

Answer

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

.

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Question

How many lines of symmetry does the above figure have?

Answer

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.

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Question

On the coordinate plane, the point is reflected about the -axis. The image is denoted . Give the distance .

Answer

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:

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Question

Consider regular Hexagon ; let and be the midpoints of $\overline{AB}$ and $\overline{DE}$ . Reflect the hexagon about $\overleftrightarrow{BE}$ , then again about $\overleftrightarrow{MN}$ . With which of the following points does the image of under these reflections coincide?

Answer

Refer to the figure below, which shows the reflection of about $\overleftrightarrow{BE}$ ; we will call this image .

Note that coincides with . Now, refer to the figure below, which shows the reflection of about $\overleftrightarrow{MN}$ ; we will call this image - the final image -

Note that coincides with , making this the correct response.

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Question

Consider regular Hexagon ; let and be the midpoints of $\overline{AB}$ and $\overline{DE}$ . Reflect the hexagon about $\overleftrightarrow{AD}$ , then again about $\overleftrightarrow{MN}$ . Which of the following clockwise rotations about the center would result in each point being its own image under this series of transformations?

Answer

Refer to the figure below, which shows the reflection of the given hexagon about $\overleftrightarrow{AD}$ ; 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 $\overleftrightarrow{MN}$ ; 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, $\frac{5}{6} \times 360 ^{\circ } = 300 ^{\circ }$ ,

is required to make each point its own image under the three transformations.

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Question

What is the result of rotating the point $(2,\frac{1}{3})$ about the origin in the plane by $180 \degree$ ?

Answer

Rotating a point

geometrically in the plane about the origin $180\degree$ is equivalent to negating the coordinates of the point algebraically to obtain

.

Thus, since our initial point was

$(2,\frac{1}{3})$

we negate both coordinates to get

$(-2,-\frac{1}{3})$

as the rotation about the origin by $180\degree$ .

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Question

Examine the figures in the above diagram. The figure at right is the result of performing which of the following transformations on the figure at left?

Answer

Examine the figure below.

If we connect the horizontal line with the line along the rotated "omega" at right, we see that it is the result of a one-sixth turn counterclockwise; the angle between them is one sixth of $360^{\circ }$ , or $\frac{1}{6} \times 360^{\circ } = 60^{\circ }$ .

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Question

Examine the figures in the above diagram. The figure at right is the result of performing which of the following transformations on the figure at left?

Answer

Examine the figure below:

If we connect the horizontal line with the line along the rotated nine at right, we see that it is the result of a one-third turn clockwise; the angle between them $\frac{1}{3} \times 360^{\circ } = 120^{\circ }$

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