Question

in exercises 15 - 18, the endpoints of $overline{cd}$ are given. find the coordinates of the mid - point $m$. (see example 1.)

1. $c(3,5)$ and $d(7,9)$
2. $c(-4,7)$ and $d(0,-3)$
3. $c(-2,0)$ and $d(4,9)$
4. $c(-8,-6)$ and $d(-4,10)$

in exercises 19 - 22, the mid - point $m$ and one endpoint of $overline{gh}$ are given. find the coordinates of the other endpoint. (see example 3.)

1. $g(5,-6)$ and $m(4,3)$ 20. $h(-3,7)$ and $m(-2,5)$

Explanation:

Step1: Recall mid - point formula

For two points $(x_1,y_1)$ and $(x_2,y_2)$, the mid - point $(x_m,y_m)$ has coordinates $x_m=\frac{x_1 + x_2}{2}$ and $y_m=\frac{y_1 + y_2}{2}$.

Step2: Solve for Exercise 15

Given $C(3,5)$ and $D(7,9)$. $x_m=\frac{3 + 7}{2}=5$, $y_m=\frac{5+9}{2}=7$. So mid - point is $(5,7)$.

Answer:

1. $(5,7)$
2. For $C(-4,7)$ and $D(0,-3)$, $x_m=\frac{-4 + 0}{2}=-2$, $y_m=\frac{7+( - 3)}{2}=2$. Answer: $(-2,2)$
3. For $C(-2,0)$ and $D(4,9)$, $x_m=\frac{-2 + 4}{2}=1$, $y_m=\frac{0 + 9}{2}=4.5$. Answer: $(1,4.5)$
4. For $C(-8,-6)$ and $D(-4,10)$, $x_m=\frac{-8+( - 4)}{2}=-6$, $y_m=\frac{-6 + 10}{2}=2$. Answer: $(-6,2)$
5. Let the other endpoint be $(x,y)$. Using mid - point formula $\frac{5 + x}{2}=4$ gives $x = 3$, $\frac{-6 + y}{2}=3$ gives $y = 12$. Answer: $(3,12)$
6. Let the other endpoint be $(x,y)$. $\frac{-3+x}{2}=-2$ gives $x=-1$, $\frac{7 + y}{2}=5$ gives $y = 3$. Answer: $(-1,3)$