Question

1. b is the midpoint of segment $overline{ac}$, and e is the midpoint of segment $overline{bd}$. you are given the following coordinates: a(-6, 3) c(4, 11) e(3, 7) find the coordinates of point d d(13, 3) 20. find the distance between the points p(3, 7) and q(-1, 2) -4, 5

Explanation:

Step1: Find coordinates of B (mid - point of AC)

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For $A(-6,3)$ and $C(4,11)$, $x_B=\frac{-6 + 4}{2}=\frac{-2}{2}=-1$, $y_B=\frac{3+11}{2}=\frac{14}{2}=7$. So $B(-1,7)$.

Step2: Use mid - point formula for BD with E as mid - point

Let the coordinates of $D$ be $(x_D,y_D)$. Since $E(3,7)$ is the mid - point of $BD$ and $B(-1,7)$, using the mid - point formula for the $x$ - coordinate: $\frac{-1+x_D}{2}=3$. Cross - multiply: $-1+x_D = 6$, then $x_D=7$. For the $y$ - coordinate: $\frac{7 + y_D}{2}=7$, cross - multiply: $7+y_D = 14$, then $y_D = 7$. So the coordinates of $D$ are $(7,7)$.

Step3: Find distance between P(3,7) and Q(-1,2)

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Here $x_1 = 3,y_1 = 7,x_2=-1,y_2 = 2$. Then $d=\sqrt{(-1 - 3)^2+(2 - 7)^2}=\sqrt{(-4)^2+(-5)^2}=\sqrt{16 + 25}=\sqrt{41}$.

Answer:

Coordinates of $D$: $(7,7)$; Distance between $P$ and $Q$: $\sqrt{41}$