Chapter 13, Problem 4RP

(a)

To determine

To find: The coordinate of the point D of the given figure.

Answer to Problem 4RP

The coordinate of the point D is $\left(10,0\right)$ .

Explanation of Solution

Given:

The given figure is shown in Figure 1

  

Figure 1

Calculation:

Consider that ABCD is the rectangle the slope of line BC and AD are equal as,

  $\begin{array}{l}\frac{y-0}{x-0}=\frac{7-7}{10-0}\\ y=0\end{array}$

Then, the value of x of the point D use the distance formula as the length BC and AD are equal then,

  $\begin{array}{l}\sqrt{{\left(x-0\right)}^2-{\left(0-0\right)}^2}=\sqrt{{\left(10-0\right)}^2+{\left(7-7\right)}^2}\\ x=10\end{array}$

Thus, the coordinate of the point D is $\left(10,0\right)$ .

(b)

To determine

To find: The coordinate of the point D of the given figure.

Answer to Problem 4RP

The coordinate of the point D is $\left(2,6\right)$ .

Explanation of Solution

Given:

The given figure is shown in Figure 1

  

Figure 1

Calculation:

Consider that ABCD is the isosceles trapezoid, here AD is parallel to BC and will have equal slope.

  $\begin{array}{l}\frac{0-0}{5-\left(-5\right)}=\frac{y-6}{x-\left(-2\right)}\\ y-6=0\\ y=6\end{array}$

Then, the value of x of the point D use the distance formula as the length DC and AB are equal then,

  $\begin{array}{l}\sqrt{{\left(x-5\right)}^2-{\left(6-0\right)}^2}=\sqrt{{\left(-2-\left(-5\right)\right)}^2+{\left(6-0\right)}^2}\\ {\left(x-5\right)}^2=9\\ \left(x-5\right)=\pm 3\\ x=2,8\end{array}$

Thus, the coordinate of the point D is $\left(2,6\right)$ .

(c)

To determine

To find: The coordinate of the point D of the given figure.

Answer to Problem 4RP

The coordinate of the point D is $\left(-4,4\right)$ .

Explanation of Solution

Given:

The given figure is shown in Figure 1

  

Figure 1

Calculation:

Consider that ABCD is the parallelogram, here DC is parallel to AB and the slopes are equal.

  $\begin{array}{l}\frac{4-y}{5-x}=\frac{-2-\left(-2\right)}{3-\left(-6\right)}\\ \frac{4-y}{5-x}=0\\ y=4\end{array}$

Consider that ABCD is the parallelogram, here AD is parallel to BC and the slopes are equal.

  $\begin{array}{l}\frac{4-\left(-2\right)}{5-\left(-6\right)}=\frac{4-\left(-2\right)}{5-3}\\ \frac{6}{x+6}=3\\ x=-4\end{array}$

Thus, the coordinate of the point D is $\left(-4,4\right)$ .