Chapter 5.5, Problem 23PSB

a

To determine

To calculate: coordinates of the point $O$ in rhombus $RHOM$

Answer to Problem 23PSB

Coordinates of $O$ are $\left(19,15\right)$

Explanation of Solution

Given information:

  $RHOM$ is rhombus, $R=\left(1,3\right),M\left(14,3\right)\text{ and }H=\left(6,15\right)$

Formula used:

• $AB$ bisects $CD$ at $O$ implies that $CO=OD$
• Midpoint of $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ is $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
• In rhombus, diagonals are perpendicular bisectors.

Calculation:

  

  $H\left(6,15\right)\text{ \& }M\left(14,3\right)$ and $T$ is the mid-point of $HM$ , implies that

  $T=\left(\frac{6+14}{2},\frac{15+3}{2}\right)=\left(\frac{20}{2},\frac{18}{2}\right)=\left(10,9\right)$

  $R\left(1,3\right)\text{ \& }T\left(10,9\right)$ and $T$ is also mid-point of $RO$ , implies that $O$ divides $RT$ externally in $2:1$

  $O=\left(\frac{2\times 10-1\times 1}{2-1},\frac{2\times 9-1\times 3}{2-1}\right)=\left(\frac{20-1}{1},\frac{18-3}{1}\right)=\left(19,15\right)$

b

To determine

To calculate: the slopes of $\overleftrightarrow{HM}\text{ and }\overleftrightarrow{RO}$

Answer to Problem 23PSB

The slope of $\overleftrightarrow{HM}$ is $-\frac{3}{2}$ , the slope of $\overleftrightarrow{RO}$ is $\frac{2}{3}$

Explanation of Solution

Given information:

  $RHOM$ is rhombus, $R=\left(1,3\right),M\left(14,3\right)\text{, }H=\left(6,15\right)\text{ and }O\left(19,15\right)$

  

Formula used:

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

Calculation:

  

The slope of $\overleftrightarrow{HM}$ is $\frac{3-15}{14-6}=\frac{-12}{8}=-\frac{3}{2}$

The slope of $\overleftrightarrow{RO}$ is $\frac{15-3}{19-1}=\frac{12}{18}=\frac{2}{3}$

c

To determine

To verify: the result in part b.

Explanation of Solution

Given information:

  $RHOM$ is rhombus, $R=\left(1,3\right),M\left(14,3\right)\text{, }H=\left(6,15\right)\text{ and }O\left(19,15\right)$ and

The slope of $\overleftrightarrow{HM}$ is $-\frac{3}{2}$ , the slope of $\overleftrightarrow{RO}$ is $\frac{2}{3}$

  

Formula used:

The product of the slopes of two perpendicular lines is $-1$ .

Calculation:

  $\left(\text{Slope of }\overleftrightarrow{HM}\right)\times \left(\text{Slope of }\overleftrightarrow{RO}\right)=\left(-\frac{3}{2}\right)\times \left(\frac{2}{3}\right)=-1$

Hence the result in part b is verified.