Question

directions: find the missing endpoint if s is the midpoint rt. 10. r(-9, 4) and s(2, -1); find t. 11. s(-4, -6) and t(-7, -3); find r. 12. b is the midpoint of ac and e is the midpoint of bd. if a(-9, -4), c(-1, 6), and e(-4, -3), find the coordinates of d.

Explanation:

Step1: Recall mid - point formula

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $M(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$.

Step2: Solve for problem 10

Let $R(x_1,y_1)=(-9,4)$ and $S(x_m,y_m)=(2,-1)$. Using the mid - point formula $x_m=\frac{x_1 + x_2}{2}$ and $y_m=\frac{y_1 + y_2}{2}$.
For the $x$ - coordinate: $2=\frac{-9 + x_2}{2}$, then $4=-9 + x_2$, so $x_2=4 + 9=13$.
For the $y$ - coordinate: $-1=\frac{4 + y_2}{2}$, then $-2=4 + y_2$, so $y_2=-2 - 4=-6$. So $T(13,-6)$.

Step3: Solve for problem 11

Let $S(x_1,y_1)=(-4,-6)$ and $T(x_2,y_2)=(-7,-3)$. Let $R(x,y)$ be the other endpoint. Using the mid - point formula $x_1=\frac{x + x_2}{2}$ and $y_1=\frac{y + y_2}{2}$.
For the $x$ - coordinate: $-4=\frac{x-7}{2}$, then $-8=x - 7$, so $x=-8 + 7=-1$.
For the $y$ - coordinate: $-6=\frac{y-3}{2}$, then $-12=y - 3$, so $y=-12 + 3=-9$. So $R(-1,-9)$.

Step4: Solve for problem 12

First, find the mid - point $B$ of $AC$. Let $A(x_1,y_1)=(-9,-4)$ and $C(x_2,y_2)=(-1,6)$.
For the $x$ - coordinate of $B$: $x_B=\frac{-9+( - 1)}{2}=\frac{-9 - 1}{2}=-5$.
For the $y$ - coordinate of $B$: $y_B=\frac{-4 + 6}{2}=1$. So $B(-5,1)$.
Let $B(x_m,y_m)=(-5,1)$ and $E(x_2,y_2)=(-4,-3)$. Let $D(x,y)$ be the other endpoint.
For the $x$ - coordinate: $-5=\frac{x-4}{2}$, then $-10=x - 4$, so $x=-10 + 4=-6$.
For the $y$ - coordinate: $1=\frac{y-3}{2}$, then $2=y - 3$, so $y=2 + 3=5$. So $D(-6,5)$.

Answer:

1. $T(13,-6)$
2. $R(-1,-9)$
3. $D(-6,5)$