Chapter 5.7, Problem 23PSB

To determine

Tofind: the slopes of the diagonals of $RHOM$

Answer to Problem 23PSB

The slopes of the diagonals $\overline{RO}\text{ and }\overline{HM}$ of $RHOM$ are of $-\frac{1}{4}\text{ and }4$ .

Explanation of Solution

Given information:

  $R,H,O,\text{ and M}$ are mid-points of the given rectangle.

Formula used:

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

Slope of the line joining $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ is $\frac{y_2-y_1}{x_2-x_1}$

Calculation:

  

Let $ABCD$ be the given rectangle.

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

We have $A\left(0,0\right)\text{, }B\left(2,8\right),C\left(18,4\right)\text{, and }D\left(16,-4\right)$

Mid-point of $\overline{AB}$ is $R=\left(\frac{0+2}{2},\frac{0+8}{2}\right)=\left(\frac{2}{2},\frac{8}{2}\right)=\left(1,4\right)$

Mid-point of $\overline{BC}$ is $H=\left(\frac{2+18}{2},\frac{8+4}{2}\right)=\left(\frac{20}{2},\frac{12}{2}\right)=\left(10,6\right)$

Mid-point of $\overline{CD}$ is $O=\left(\frac{18+16}{2},\frac{4-4}{2}\right)=\left(\frac{34}{2},\frac{0}{2}\right)=\left(17,0\right)$

Mid-point of $\overline{DA}$ is $M=\left(\frac{16+0}{2},\frac{-4+0}{2}\right)=\left(\frac{16}{2},\frac{-4}{2}\right)=\left(8,-2\right)$

Now, we have $R=\left(1,4\right),H=\left(10,6\right),O=\left(17,0\right),M=\left(8,-2\right)$

The diagonals of $RHOM$ are $\overline{RO}\text{ and }\overline{HM}$

The slope of $\overline{RO}$ is $\frac{0-4}{17-1}=\frac{-4}{16}=-\frac{1}{4}$

The slope of $\overline{HM}$ is $\frac{-2-6}{8-10}=\frac{-8}{-2}=4$