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1) graph the points a (1, 6) and b (9, 6). find the mid - point of $ove…

Question

  1. graph the points a (1, 6) and b (9, 6). find the mid - point of $overline{ab}$. find the distance of $overline{ab}$
  2. graph the points c (2, 2) and d (6, 2). find the mid - point of $overline{cd}$. find the distance of $overline{cd}$
  3. graph the points e (-10, -9) and f (-10, -3). find the mid - point of $overline{ef}$. find the distance $overline{ef}$

Explanation:

Response
1. For points A(1, 6) and B(9, 6)

Step1: Find the 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: Calculate the mid - point of AB

Here $x_1 = 1,y_1=6,x_2 = 9,y_2 = 6$. So, $M_{AB}=(\frac{1 + 9}{2},\frac{6+6}{2})=(5,6)$.

Step3: Find the distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step4: Calculate the distance of AB

Substitute $x_1 = 1,y_1=6,x_2 = 9,y_2 = 6$ into the formula: $d_{AB}=\sqrt{(9 - 1)^2+(6 - 6)^2}=\sqrt{8^2+0^2}=8$.

2. For points C(2, 2) and D(6, 2)

Step1: Calculate the mid - point of CD

Using the mid - point formula with $x_1 = 2,y_1=2,x_2 = 6,y_2 = 2$, we get $M_{CD}=(\frac{2 + 6}{2},\frac{2+2}{2})=(4,2)$.

Step2: Calculate the distance of CD

Using the distance formula with $x_1 = 2,y_1=2,x_2 = 6,y_2 = 2$, we have $d_{CD}=\sqrt{(6 - 2)^2+(2 - 2)^2}=\sqrt{4^2+0^2}=4$.

3. For points E(- 10,-9) and F(-10,-3)

Step1: Calculate the mid - point of EF

Using the mid - point formula with $x_1=-10,y_1=-9,x_2=-10,y_2=-3$, we get $M_{EF}=(\frac{-10-10}{2},\frac{-9 - 3}{2})=(-10,-6)$.

Step2: Calculate the distance of EF

Using the distance formula with $x_1=-10,y_1=-9,x_2=-10,y_2=-3$, we have $d_{EF}=\sqrt{(-10 + 10)^2+(-3+9)^2}=\sqrt{0^2+6^2}=6$.

Answer:

  1. Mid - point of $\overline{AB}$: $(5,6)$, Distance of $\overline{AB}$: $8$
  2. Mid - point of $\overline{CD}$: $(4,2)$, Distance of $\overline{CD}$: $4$
  3. Mid - point of $\overline{EF}$: $(-10,-6)$, Distance of $\overline{EF}$: $6$