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.

If the graph of an equation is translated to the right \(\displaystyle h\) units, and upward \(\displaystyle k\) units, the equation of the image can be found by replacing \(\displaystyle x\) with \(\displaystyle x-h\) and \(\displaystyle y\) with \(\displaystyle y-k\) in the equation of the original graph. 

Since we are moving the graph of the equation

\(\displaystyle 4x- 7y = 15\)

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

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

Simplify by distributing:

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

\(\displaystyle 4 x+16- 7 y-42 = 15\)

Collect like terms:

\(\displaystyle 4 x- 7 y +16-42 = 15\)

\(\displaystyle 4 x- 7 y -26 = 15\)

Add 26 to both sides:

\(\displaystyle 4 x- 7 y -26 +26 = 15+26\)

\(\displaystyle 4 x- 7 y = 41\),

the correct choice.

If the graph of an equation is translated to the right \(\displaystyle h\) units, and upward \(\displaystyle k\) units, the equation of the image can be found by replacing \(\displaystyle x\) with \(\displaystyle x-h\) and \(\displaystyle y\) with \(\displaystyle y-k\) in the equation of the original graph. 

Since we are moving the graph of the equation

\(\displaystyle x^{2}+ y^{2}= 100\)

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

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

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

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

\(\displaystyle x^{2}- 4x+4+ y^{2}-10y+2 5 = 100\)

Collect like terms; the equation becomes

\(\displaystyle x^{2}- 4x + y^{2}-10y+29 = 100\)

Subtract 100 from both sides:

\(\displaystyle x^{2}- 4x + y^{2}-10y+29- 100 = 100 - 100\)

\(\displaystyle x^{2}- 4x + y^{2}-10y-71=0\)

If the graph of an equation is translated to the right \(\displaystyle h\) units, and upward \(\displaystyle k\) units, the equation of the image can be found by replacing \(\displaystyle x\) with \(\displaystyle x-h\) and \(\displaystyle y\) with \(\displaystyle y-k\) in the equation of the original graph. 

Since we are moving the graph of the equation

\(\displaystyle y = x^{2}+ 5x - 7\)

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

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

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

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

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

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

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

\(\displaystyle y+2 = x^{2} -8x+16+ 5x-20 - 7\)

Collect like terms on the right:

\(\displaystyle y+2 = x^{2} -8x+ 5x+16 -20 - 7\)

\(\displaystyle y+2 = x^{2} -3x-11\)

Subtract 2 from both sides:

\(\displaystyle y+2 -2= x^{2} -3x-11-2\)

\(\displaystyle y= x^{2} -3x-13\),

the equation of the image.

If the graph of an equation is translated to the right \(\displaystyle h\) units, and upward \(\displaystyle k\) units, the equation of the image can be found by replacing \(\displaystyle x\) with \(\displaystyle x-h\) and \(\displaystyle y\) with \(\displaystyle y-k\) in the equation of the original graph. 

Since we are moving the graph of the equation

\(\displaystyle y = | 3x - 8 |\)

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

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

We can simplify the expression on the right by distributing:

\(\displaystyle y+5 = | 3x+3 \cdot 3 - 8 |\)

\(\displaystyle y+5 = | 3x+9- 8 |\)

Collect like terms:

\(\displaystyle y+5 = | 3x+1 |\)

Subtract 5 from both sides:

\(\displaystyle y+5 -5 = | 3x+1 | - 5\)

\(\displaystyle y = | 3x+1 | - 5\),

the correct equation of the image.

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

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

Substitute the coordinates of \(\displaystyle A\) and \(\displaystyle B\) in this formula to find that midpoint \(\displaystyle M\) of \(\displaystyle \overline{AB}\) is located at

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

Substitute the coordinates of \(\displaystyle A\) and \(\displaystyle M\) to find that midpoint \(\displaystyle N\) of \(\displaystyle \overline{AM}\) is located at

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

To perform the translation \(\displaystyle N \rightarrow B\), or, equivalently,

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

on a point, it is necessary to add

\(\displaystyle 9 - \frac{9}{2} = \frac{18}{2} - \frac{9}{2} = \frac{9}{2}\)

to its \(\displaystyle x\)-coordinate, and

\(\displaystyle -5 - \frac{11}{2} = - \frac{10}{2} - \frac{11}{2} = - \frac{21}{2}\)

to its \(\displaystyle y\)-coordinate.

Therefore, the \(\displaystyle x\)-coordinate of the image of \(\displaystyle M\) under this translation is

\(\displaystyle 6+ \frac{9}{2} = \frac{12}{2}+ \frac{9}{2} = \frac{21}{2}\);

its \(\displaystyle y\)-coordinate is

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

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

The translation \(\displaystyle A\rightarrow B\) on a figure is the translation that shifts a figure so that the image of \(\displaystyle A\), which we will call \(\displaystyle A''\), coincides with \(\displaystyle B\). 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 \(\displaystyle F\) marked as \(\displaystyle F''\):

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

\(\displaystyle F'\), the image of \(\displaystyle F\) under these two transformations, coincides with \(\displaystyle C\).

The translation \(\displaystyle A\rightarrow B\) on a figure is the translation that shifts a figure so that the image of \(\displaystyle A\), which we will call \(\displaystyle A''\), coincides with \(\displaystyle B\). 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 \(\displaystyle 120 ^{\circ }\) - one third of a turn - about \(\displaystyle B\), and call \(\displaystyle A '\) the image of \(\displaystyle A''\), and so forth, the result is as follows:

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

The translation \(\displaystyle F\rightarrow C\) on a figure is the translation that shifts a figure so that the image of \(\displaystyle F\) coincides with \(\displaystyle C\). All other points shift the same distance in the same direction. Below shows the image of the given hexagon under this translation, with \(\displaystyle D''\) the image of \(\displaystyle D\):

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

\(\displaystyle D'\) coincides with \(\displaystyle A\) in the original hexagon, making \(\displaystyle A\) the correct response.

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 \(\displaystyle A\) are \(\displaystyle A\) itself, which is at \(\displaystyle (8, 0)\), and \(\displaystyle M\), which itself is the midpoint of the side with origin \(\displaystyle O\) and \(\displaystyle B\), which is \(\displaystyle (6, 4)\), as its endpoints.

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

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

so, substituting the coordinates of \(\displaystyle O\) and \(\displaystyle B\) in the formula, we see that \(\displaystyle M\) is

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

See the figure below:

To perform the translation \(\displaystyle A \rightarrow M\), or, equivalently,

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

on a point, it is necessary to add

\(\displaystyle 3-8 = -5\)

and

\(\displaystyle 2- 0 = 2\)

to the \(\displaystyle x\)- and \(\displaystyle y\)- coordinates, respectively. Therefore, the image of \(\displaystyle B\) is located at 

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

or

\(\displaystyle (1,6)\).