## Geometry: Common Core (15th Edition)

Use the distance formula to find the length of $\overline{AB}$. $d_{AB}=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$ substitute $d_{AB}=\sqrt{(3-(-4))^2+(5-5)^2}$ $d_{AB}=\sqrt{(7)^2+(0)^2}$ $d_{AB}=\sqrt{49+0}$ $d_{AB}=\sqrt{49}$ $d_{AB}=7$ Since the y-coordinate of A and B are the same, $\overline{AB}$ is a horizontal line. If B is the midpoint of $\overline{AC}$ then A, B and C are collinear and $\overline{AC}$ is also a horizontal line with a y-coordinate of 5. If B is the midpoint of $\overline{AC}$, the length of $\overline{AC}$ is twice the length of $\overline{AB}$. $AC=2(AB)=2(7)=14$ Use the distance formula to find the x-coordinate of point C. $d_{AC}=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}$ substitute $14=\sqrt{(x_C-(-4))^2+(5-5)^2}$ $14=\sqrt{(x_C+4)^2+(0)^2}$ $14=\sqrt{(x_C+4)^2}$ $14=x_C+4$ subtract 4 from each side $14-4=x_C+4-4$ $10=x_C$