The image of a point $\displaystyle P$ under a dilation with center $\displaystyle O$ and scale factor $\displaystyle s$ is the point in the same direction from $\displaystyle O$ as , but $\displaystyle s \cdot OP$ units away, where $\displaystyle OP$ is the distance from $\displaystyle O$ to $\displaystyle P$. Since the center of the dilation is the origin $\displaystyle (0,0)$, the coordinates of the image of $\displaystyle A$, which is $\displaystyle A'$, can be found by multiplying scale factor $\displaystyle \frac{2}{5}$ by each of those of $\displaystyle A$:

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

$\displaystyle A':$

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

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

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

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

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

$\displaystyle B' :$

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

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

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

$\displaystyle C':$

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

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

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

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

$\displaystyle D :$

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

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

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

To perform a dilation with center $\displaystyle A$ and scale factor $\displaystyle \frac{1}{2}$, find the midpoints of the segments connecting $\displaystyle A$ to each point, and connect those points. We can simplify the process by finding the midpoints of $\displaystyle \overline{AB}$, $\displaystyle \overline{AC}$, and $\displaystyle \overline{AD}$, and naming them $\displaystyle B''$, $\displaystyle C''$, and $\displaystyle D''$, respectively; $\displaystyle A''$, the image of center $\displaystyle A$, is just $\displaystyle A$ itself. The figure is below:

Now, do the same thing to the new square, but with $\displaystyle B$ as the center. The figure is below:

The final image, relative to the original square, is below:

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  $\displaystyle O$:

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

Removing the medians and $\displaystyle O$, we see that the correct choice is the figure

The graph of the equation

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

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

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

or

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

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

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 $\displaystyle \frac{1}{3}$ 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 $\displaystyle \frac{1}{3}$; the endpoints of the horizontal axis will be

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

or

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

and the endpoints of the vertical axis will be

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

or

$\displaystyle (0, \pm 2)$.

This dilation is shown in the figure below:

Therefore, $\displaystyle a = 3$ and $\displaystyle b = 2$, making the equation of the image

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

or

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

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, $\displaystyle \overline{AB }$ and $\displaystyle \overline{CB}$ are two of the altitudes, which intersect at $\displaystyle B$; the third altitude must also pass through $\displaystyle B$, since the three altitudes are concurrent. Therefore, we perform a dilation of the triangle with respect to center $\displaystyle B$.

This is done by mapping $\displaystyle A$ and $\displaystyle C$ to the midpoints of $\displaystyle \overline{AB }$ and $\displaystyle \overline{CB}$, respectively, and by mapping $\displaystyle B$ to itself. The triangle is seen below:

This figure is the correct choice.

To perform a dilation with center $\displaystyle C$ and scale factor $\displaystyle \frac{1}{2}$, find the midpoints of the segments connecting $\displaystyle C$ to each point, and connect those points. We can simplify the process by finding the midpoints of $\displaystyle \overline{CA}$, $\displaystyle \overline{CB}$, and $\displaystyle \overline{CD}$, and naming them $\displaystyle A''$, $\displaystyle B''$, and $\displaystyle D''$, respectively; $\displaystyle C''$, the image of center $\displaystyle C$, is just $\displaystyle C$ itself. The figure is below:

Now, do the same thing to the new square, but with $\displaystyle A$ as the center. The figure is below:

The image, relative to the original square, is below:

The graph of the equation

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

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

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

or

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

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

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

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

or

$\displaystyle (\pm 12,0)$,

and the endpoints of the vertical axis will be

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

or

$\displaystyle (0, \pm 9)$.

The figure is shown below:

Therefore, $\displaystyle a = 12$ and $\displaystyle b = 9$, making the equation of the image

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

or

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

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 $\displaystyle \overline{AB}$ and $\displaystyle \overline{AC}$ to $\displaystyle \overleftrightarrow{AB}$ and $\displaystyle \overleftrightarrow{AC}$;  the altitudes are shown below, with point of intersection  $\displaystyle O$:

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

Removing the altitudes and $\displaystyle O$, we see that the correct choice is the figure:

The graph of the equation

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

is a circle with center $\displaystyle (h, k)$ and radius $\displaystyle r$.

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

or

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

is a circle with  center at origin $\displaystyle (0,0)$ and radius 8.

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

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

Substituting in the circle formula, this is 

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

or

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

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  $\displaystyle O$:

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

Removing the perpendicular bisectors and $\displaystyle O$, we see that the correct choice is the figure