Question

in exercises 23 - 26, the endpoints of $overline{cd}$ are given. find the coordinates of the midpoint $m$. check your answer. (see example 5.) 23. $c(3, - 5)$ and $d(7,9)$ 24. $c(-4,7)$ and $d(0, - 3)$ 25. $c(-2,0)$ and $d(4,9)$ 26. $c(-8, - 6)$ and $d(-4,10)$

Explanation:

Step1: Recall mid - point formula

The mid - point formula for 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: For point 23

Given $C(3,-5)$ and $D(7,9)$, where $x_1 = 3,y_1=-5,x_2 = 7,y_2 = 9$. Then $x$ - coordinate of mid - point $M_x=\frac{3 + 7}{2}=\frac{10}{2}=5$, and $y$ - coordinate of mid - point $M_y=\frac{-5 + 9}{2}=\frac{4}{2}=2$. So the mid - point $M(5,2)$.

Step3: For point 24

Given $C(-4,7)$ and $D(0,-3)$, where $x_1=-4,y_1 = 7,x_2 = 0,y_2=-3$. Then $x$ - coordinate of mid - point $M_x=\frac{-4+0}{2}=\frac{-4}{2}=-2$, and $y$ - coordinate of mid - point $M_y=\frac{7+( - 3)}{2}=\frac{4}{2}=2$. So the mid - point $M(-2,2)$.

Step4: For point 25

Given $C(-2,0)$ and $D(4,9)$, where $x_1=-2,y_1 = 0,x_2 = 4,y_2 = 9$. Then $x$ - coordinate of mid - point $M_x=\frac{-2 + 4}{2}=\frac{2}{2}=1$, and $y$ - coordinate of mid - point $M_y=\frac{0 + 9}{2}=4.5$. So the mid - point $M(1,4.5)$.

Step5: For point 26

Given $C(-8,-6)$ and $D(-4,10)$, where $x_1=-8,y_1=-6,x_2=-4,y_2 = 10$. Then $x$ - coordinate of mid - point $M_x=\frac{-8+( - 4)}{2}=\frac{-12}{2}=-6$, and $y$ - coordinate of mid - point $M_y=\frac{-6 + 10}{2}=\frac{4}{2}=2$. So the mid - point $M(-6,2)$.

Answer:

1. $M(5,2)$
2. $M(-2,2)$
3. $M(1,4.5)$
4. $M(-6,2)$