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HiSET: Math : Translations

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Example Questions

Example Question #1 : Translations

Translation

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?

Possible Answers:

Reflection in the y-axis

Rotation about the origin

Reflection in the x-axis

Dilation

Translation

Correct answer:

Translation

Explanation:

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.

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Example Question #2 : Translations

Translate the graph of the equation

4x- 7y = 15

left four units and down six units. Give the equation of the image.

Possible Answers:

4x- 7y = 45

4x- 7y = 25

4 x- 7 y = −43

4 x- 7 y = 73

4 x- 7 y = 41

Correct answer:

4 x- 7 y = 41

Explanation:

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 x-h and with y-k in the equation of the original graph.

Since we are moving the graph of the equation

4x- 7y = 15

left four units and down six units, we set h = -4 and k = -6 ; we can replace with x - (-4) , or x+ 4 , and with y - (-6 ) , or y + 6 . The equation of the image can be written as

4 (x+ 4)- 7 (y+6) = 15

Simplify by distributing:

4 x+4( 4)- 7 y-7(6) = 15

4 x+16- 7 y-42 = 15

Collect like terms:

4 x- 7 y +16-42 = 15

4 x- 7 y -26 = 15

Add 26 to both sides:

4 x- 7 y -26 +26 = 15+26

4 x- 7 y = 41 ,

the correct choice.

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Example Question #6 : Understand Transformations In The Plane

Translate the graph of the equation

$x^{2}+ y^{2}= 100$

right two units and up five units. Give the equation of the image.

Possible Answers:

None of the other choices gives the correct response.

$x^{2}+ 4x + y^{2}+10y-71=0$

$x^{2}- 4x + y^{2}-10y-71=0$

$x^{2}+ 2x + y^{2}+5y-71=0$

$x^{2}- 2x + y^{2}-5y-71=0$

Correct answer:

$x^{2}- 4x + y^{2}-10y-71=0$

Explanation:

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 x-h and with y-k in the equation of the original graph.

Since we are moving the graph of the equation

$x^{2}+ y^{2}= 100$

right two units and up five units, we set h = 2 and k = 5 ; we can therefore replace with x- 2 and with y - 5 . The equation of the image can be written as

$(x-2)^{2}+( y-5)^{2}= 100$

This can be rewritten by applying the binomial square pattern as follows:

$x^{2}- 2 (x) (2)+ 2^{2}+ y^{2}- 2 (5)(x)+ 5^{2} = 100$

$x^{2}- 4x+4+ y^{2}-10y+2 5 = 100$

Collect like terms; the equation becomes

$x^{2}- 4x + y^{2}-10y+29 = 100$

Subtract 100 from both sides:

$x^{2}- 4x + y^{2}-10y+29- 100 = 100 - 100$

$x^{2}- 4x + y^{2}-10y-71=0$

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Example Question #102 : Hi Set: High School Equivalency Test: Math

Translate the graph of the equation

$y = x^{2}+ 5x - 7$

right four units and down two units. Give the equation of the image.

Possible Answers:

$y = x^{2}+ 13x -9$

$y = x^{2}+ 13x + 29$

$y= x^{2} -3x-13$

$y= x^{2} -3x-9$

$y = x^{2}+ 13x + 27$

Correct answer:

$y= x^{2} -3x-13$

Explanation:

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 x-h and with y-k in the equation of the original graph.

Since we are moving the graph of the equation

$y = x^{2}+ 5x - 7$

right four units and down two units, we set h = 4 and k = -2 ; we can therefore replace with x- 4 , and with y - (-2) , or y+ 2 . The equation of the image can be written as

$y+2 = (x-4)^{2}+ 5(x-4) - 7$

The expression at right can be simplified. First, use the distributive property on the middle expression:

$y+2 = (x-4)^{2}+ 5x-5 (4) - 7$

$y+2 = (x-4)^{2}+ 5x-20 - 7$

Now, simplify the first expression by using the binomial square pattern:

$y+2 = x^{2} -2 (x)(4) +4^{2}+ 5x-20 - 7$

$y+2 = x^{2} -8x+16+ 5x-20 - 7$

Collect like terms on the right:

$y+2 = x^{2} -8x+ 5x+16 -20 - 7$

$y+2 = x^{2} -3x-11$

Subtract 2 from both sides:

$y+2 -2= x^{2} -3x-11-2$

$y= x^{2} -3x-13$ ,

the equation of the image.

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Example Question #103 : Hi Set: High School Equivalency Test: Math

Translate the graph of the equation

y = | 3x - 8 |

left three units and down five units. Give the equation of the image.

Possible Answers:

y = | 3x-17 | + 5

y = | 3x-17 | - 5

None of the other choices gives the correct response.

y = | 3x+1 | - 5

y = | 3x+1 | +5

Correct answer:

y = | 3x+1 | - 5

Explanation:

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 x-h and with y-k in the equation of the original graph.

Since we are moving the graph of the equation

y = | 3x - 8 |

left three units and down five units, we set h = -3 and k = -5 ; we can therefore replace with x- (-3) , or x +3 , and with y - (-5) , or y+ 5 . The equation of the image can be written as

y+5 = | 3(x+3) - 8 |

We can simplify the expression on the right by distributing:

$y+5 = | 3x+3 \cdot 3 - 8 |$

y+5 = | 3x+9- 8 |

Collect like terms:

y+5 = | 3x+1 |

Subtract 5 from both sides:

y+5 -5 = | 3x+1 | - 5

y = | 3x+1 | - 5 ,

the correct equation of the image.

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Example Question #104 : Hi Set: High School Equivalency Test: Math

On the coordinate plane, let and be located at (3, 9) and (9, -5) , respectively. Let be the midpoint of $\overline{AB}$ and let be the midpoint of $\overline{AM}$ . On the segment, perform the translation $N \rightarrow B$ . Where is the image of located?

Possible Answers:

(6, 2)

$\left ( \frac{15}{2}, -\frac{3}{2} \right )$

$\left ( \frac{25}{2}, \frac{3}{2} \right )$

(12, -12)

$\left (\frac{21}{2},-\frac{17}{2} \right )$

Correct answer:

$\left (\frac{21}{2},-\frac{17}{2} \right )$

Explanation:

The midpoint of a segment with endpoints at $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is located at

$\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$

Substitute the coordinates of and in this formula to find that midpoint of $\overline{AB}$ is located at

$\left ( \frac{3+9}{2}, \frac{9+(-5)}{2} \right )$ , or (6, 2) .

Substitute the coordinates of and to find that midpoint of $\overline{AM}$ is located at

$\left ( \frac{3+6}{2}, \frac{9+2}{2} \right )$ , or $\left ( \frac{9}{2}, \frac{11}{2} \right )$ .

To perform the translation $N \rightarrow B$ , or, equivalently,

$\left ( \frac{9}{2}, \frac{11}{2} \right ) \rightarrow (9, -5)$ ,

on a point, it is necessary to add

$9 - \frac{9}{2} = \frac{18}{2} - \frac{9}{2} = \frac{9}{2}$

to its -coordinate, and

$-5 - \frac{11}{2} = - \frac{10}{2} - \frac{11}{2} = - \frac{21}{2}$

to its -coordinate.

Therefore, the -coordinate of the image of under this translation is

$6+ \frac{9}{2} = \frac{12}{2}+ \frac{9}{2} = \frac{21}{2}$ ;

its -coordinate is

$2 + \left (- \frac{21}{2} \right ) = \frac{4}{2}- \frac{21}{2}=-\frac{17}{2}$

The image of is located at $\left (\frac{21}{2},-\frac{17}{2} \right )$ .

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Example Question #2 : Translations

Consider regular Hexagon ABCDEF .

On this hexagon, perform the translation $A\rightarrow B$ . Then reflect the hexagon about $\overleftrightarrow{BD}$ . Let be the image of under these transformations, and so forth.

Which point on Hexagon ABCDEF is the image of under these transformations?

Possible Answers:

Correct answer:

Explanation:

The translation $A\rightarrow B$ on a figure is the translation that shifts a figure so that the image of , which we will call A'' , 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 F'' :

If the image is reflected about $\overleftrightarrow{BD}$ , the new image is the original hexagon. Calling the image of F'' under this reflection, we get the following:

, the image of under these two transformations, coincides with .

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Example Question #7 : Translations

Consider regular Hexagon ABCDEF .

On this hexagon, perform the translation $A\rightarrow B$ . Then perform a $120^{\circ }$ 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 A'B'C'D'E'F' relative to Hexagon ABCDEF ?

Possible Answers:

Hexagons

Correct answer:

Hexagons

Explanation:

If this new hexagon is rotated clockwise $120 ^{\circ }$ - one third of a turn - about , and call A ' the image of A'' , and so forth, the result is as follows:

Removing the intermediate markings, we see that the correct response is

Hexagons

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Example Question #111 : Hi Set: High School Equivalency Test: Math

Consider regular Hexagon ABCDEF .

On this hexagon, perform the translation $F\rightarrow C$ . Then perform a $180^{\circ }$ 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?

Possible Answers:

Correct answer:

Explanation:

The translation $F\rightarrow C$ 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 D'' the image of :

If this new hexagon is rotated $180 ^{\circ }$ - one half of a turn - about - the image is the original hexagon, but the vertices can be relabeled. Letting be the image of D'' under this rotation, and so forth:

coincides with in the original hexagon, making the correct response.

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Example Question #2 : Translations

On the coordinate plane, let , , and be located at the origin, (8, 0) , and (6, 4) . Construct the median of $\bigtriangleup OAB$ from and let the foot of the median be . On the triangle, perform the translation $A \rightarrow M$ . Where is the image of ?

Possible Answers:

(11,2)

(10,4)

(-3, -2)

(9,6)

(1,6)

Correct answer:

(1,6)

Explanation:

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 (8, 0) , and , which itself is the midpoint of the side with origin and , which is (6, 4) , as its endpoints.

The midpoint of a segment with endpoints at $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is located at

$\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$ ,

so, substituting the coordinates of and in the formula, we see that is

$\left ( \frac{0+6}{2}, \frac{0+4}{2} \right )$ , or (3,2) .

See the figure below:

To perform the translation $A \rightarrow M$ , or, equivalently,

$(8, 0) \rightarrow (3,2)$ ,

on a point, it is necessary to add

3-8 = -5

and

2- 0 = 2

to the - and - coordinates, respectively. Therefore, the image of is located at

(6+ (-5), 4+ 2) ,

(1,6) .

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HiSET: Math : Translations

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #1 : Translations

Example Question #2 : Translations

Example Question #6 : Understand Transformations In The Plane

Example Question #102 : Hi Set: High School Equivalency Test: Math

Example Question #103 : Hi Set: High School Equivalency Test: Math

Example Question #104 : Hi Set: High School Equivalency Test: Math

Example Question #2 : Translations

Example Question #7 : Translations

Example Question #111 : Hi Set: High School Equivalency Test: Math

Example Question #2 : Translations

All HiSET: Math Resources

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