To find the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$, use the distance formula: $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$. Here, the points are $(-3,4)$ and $(5,-2)$, so substitute carefully: $\sqrt{(5-(-3){)}^2+(-2-4{)}^2}$. That matches choice A.

To find the midpoint of a segment, average the $x$-coordinates and average the $y$-coordinates. For the points $(2,-1)$ and $(8,5)$, the midpoint is

So the correct answer is $(5,2)$.

The first line has slope $-3$. A line perpendicular to it has slope equal to the negative reciprocal of $-3$, which is $\frac{1}{3}$. Since the second line passes through $(6,1)$, use point-slope form: $y-1=\frac{1}{3}(x-6)$. So the correct answer is $y-1=\frac{1}{3}(x-6)$.

Use the distance formula between $(x_1,y_1)$ and $(x_2,y_2)$:

For points $P(2,5)$ and $Q(8,1)$, the change in $x$ is $8-2=6$ and the change in $y$ is $1-5=-4$. So,

Therefore, the distance between $P$ and $Q$ is $\sqrt{52}$ grid units.

To find the midpoint of a segment with endpoints $(x_1,y_1)$ and $(x_2,y_2)$, use the midpoint formula:

For $A(2,5)$ and $B(8,1)$, the midpoint is

So the statement that must be true is that the midpoint of $\overline{AB}$ is $(5,3)$.