HSG.GPE.B.6 math practice. Learn by doing.

Let $B = (x, y)\textsf{.}$ Since $M = (4, 7)$ is the midpoint of $A = (1, 5)$ and $B\textsf{,}$ the midpoint formula gives $\dfrac{1 + x}{2} = 4$ and $\dfrac{5 + y}{2} = 7\textsf{.}$ Multiply $\dfrac{1 + x}{2} = 4$ by $2$ to get $1 + x = 8\textsf{,}$ so $x = 7\textsf{.}$ Multiply $\dfrac{5 + y}{2} = 7$ by $2$ to get $5 + y = 14\textsf{,}$ so $y = 9\textsf{.}$ The coordinates of $B$ are $(7, 9)\textsf{.}$

The midpoint averages the matching coordinates: $\left(\dfrac{2 + 8}{2}\textsf{,} \dfrac{5 + 1}{2}\right) = \left(\dfrac{10}{2}\textsf{,} \dfrac{6}{2}\right) = (5, 3)\textsf{.}$

$x$-coordinate $= \dfrac{2 + 6}{2} = \dfrac{8}{2} = 4$.

$y$-coordinate $= \dfrac{8 + 4}{2} = \dfrac{12}{2} = 6$.

Let $P = (x, y)\textsf{.}$ Since $(6, 2)$ is the midpoint of $P$ and $Q = (9, 8)\textsf{,}$ the midpoint formula gives $\dfrac{x + 9}{2} = 6$ and $\dfrac{y + 8}{2} = 2\textsf{.}$ Multiply $\dfrac{x + 9}{2} = 6$ by $2$ to get $x + 9 = 12\textsf{,}$ so $x = 3\textsf{.}$ Multiply $\dfrac{y + 8}{2} = 2$ by $2$ to get $y + 8 = 4\textsf{,}$ so $y = -4\textsf{.}$ The coordinates of $P$ are $(3, -4)\textsf{.}$

The midpoint averages the matching coordinates: $\left(\dfrac{3 + 9}{2}\textsf{,} \dfrac{7 + 11}{2}\right) = \left(\dfrac{12}{2}\textsf{,} \dfrac{18}{2}\right) = (6, 9)\textsf{.}$

$x$-coordinate $= \dfrac{0 + 8}{2} = \dfrac{8}{2} = 4$.

$y$-coordinate $= \dfrac{10 + 2}{2} = \dfrac{12}{2} = 6$.