Chapter 5.7, Problem 22PSB

a

To determine

To find: the coordinates of $M,N,O,\text{ and }P$ .

Answer to Problem 22PSB

The coordinates of $M,N,O,\text{ and }P$ are $M=\left(2,4\right)\text{, }N=\left(4,8\right)\text{, }O=\left(6,6\right),\text{ and }P=\left(4,2\right)$

Explanation of Solution

Given information:

  $M,N,O,\text{ and }P$ are mid-points of the sides of $ABCD$

Formula used:

Mid-point of $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ is $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Calculation:

  

Let the coordinates be $A\left(1,1\right),B\left(3,7\right),C\left(5,9\right),D\left(7,3\right)$

Given that $M,N,O,\text{ and }P$ are mid-points of $\overline{AB,}\overline{BC}\text{, }\overline{CD}\text{, and }\overline{DA}$ respectively.

  $M=\left(\frac{1+3}{2},\frac{1+7}{2}\right)=\left(\frac{4}{2},\frac{8}{2}\right)=\left(2,4\right)$

  $N=\left(\frac{3+5}{2},\frac{7+9}{2}\right)=\left(\frac{8}{2},\frac{16}{2}\right)=\left(4,8\right)$

  $O=\left(\frac{5+7}{2},\frac{9+3}{2}\right)=\left(\frac{12}{2},\frac{12}{2}\right)=\left(6,6\right)$

  $P=\left(\frac{7+1}{2},\frac{3+1}{2}\right)=\left(\frac{8}{2},\frac{4}{2}\right)=\left(4,2\right)$

b

To determine

To calculate: the slopes of $\overleftrightarrow{MN}\text{ and }\overleftrightarrow{PO}$

Answer to Problem 22PSB

The slopes of $\overleftrightarrow{MN}\text{ and }\overleftrightarrow{OP}$ are, $m_1=2\text{ and }{m}_2=2$

Explanation of Solution

Given information:

  $M,N,O,\text{ and }P$ are mid-points of the sides of $ABCD$ and

  $M=\left(2,4\right)\text{, }N=\left(4,8\right)\text{, }O=\left(6,6\right),\text{ and }P=\left(4,2\right)$

Formula used:

The slope of $\left(x_1,y_1\right)\text{ and}\left(x_2,y_2\right)$ is, $m=\frac{y_2-y_1}{x_2-x_1}$

Calculation:

  

The slope of $M\left(2,4\right)\text{ and N}\left(4,8\right)$ is, $m_1=\frac{8-4}{4-2}=\frac{4}{2}=2$

The slope of $O\left(6,6\right)\text{ and P}\left(4,2\right)$ is, $m_2=\frac{2-6}{4-6}=\frac{-4}{-2}=2$

c

To determine

To identify: true about $\overleftrightarrow{MN}\text{ and }\overleftrightarrow{OP}$ .

Answer to Problem 22PSB

Since both slopes of $\overleftrightarrow{MN}\text{ and }\overleftrightarrow{OP}$ are same, both are parallel.

Explanation of Solution

Given information:

  $M,N,O,\text{ and }P$ are mid-points of the sides of $ABCD$ and

  $M=\left(2,4\right)\text{, }N=\left(4,8\right)\text{, }O=\left(6,6\right),\text{ and }P=\left(4,2\right)$

The slopes of $\overleftrightarrow{MN}\text{ and }\overleftrightarrow{OP}$ are, $m_1=2\text{ and }{m}_2=2$

The slope of $\overleftrightarrow{MN}$ = The slope of $\overleftrightarrow{OP}$

Implies that $\overleftrightarrow{MN}\parallel \overleftrightarrow{OP}$