### All HiSET: Math Resources

## Example Questions

### Example Question #1 : Translations

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

Reflection in the x-axis

Translation

Rotation about the origin

Dilation

**Correct answer:**

Translation

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.

### Example Question #2 : Translations

Translate the graph of the equation

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

**Possible Answers:**

**Correct answer:**

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.

### Example Question #3 : Translations

Translate the graph of the equation

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

**Possible Answers:**

None of the other choices gives the correct response.

**Correct answer:**

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:

### Example Question #4 : Translations

Translate the graph of the equation

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

**Possible Answers:**

**Correct answer:**

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.

### Example Question #5 : Translations

Translate the graph of the equation

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

**Possible Answers:**

None of the other choices gives the correct response.

**Correct answer:**

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.

### Example Question #6 : Translations

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?

**Possible Answers:**

**Correct answer:**

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 .

### Example Question #7 : Translations

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?

**Possible Answers:**

**Correct answer:**

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 .

### Example Question #8 : Translations

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 ?

**Possible Answers:**

**Correct answer:**

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

### Example Question #9 : Translations

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?

**Possible Answers:**

**Correct answer:**

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.

### Example Question #10 : Translations

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 ?

**Possible Answers:**

**Correct answer:**

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

.