Chapter 13.1, Problem 8CE

a.

To determine

To find the coordinates of T

Answer to Problem 8CE

The value of T is $\left(-3,-3\right)$

Explanation of Solution

Looking at the graph given in the question

The given point is

  $\left(3,-3\right)\text{ and }\left(-3,3\right)$

Calculation:

The coordinates of T can be obtained as

The point T lies on the horizontal line of $\left(3,-3\right)$ and vertical line of $\left(-3,3\right)$

Let, $\left(x_1,y_1\right)\text{ be }\left(3,-3\right)\text{ and }\left(x_2,y_2\right)\text{ be }\left(-3,3\right)$

Then the point T is $\left(x_2,y_1\right)=\left(-3,-3\right)$

Hence,

The value of T is $\left(-3,-3\right)$

b.

To determine

To find the length of the legs of the right triangle

Answer to Problem 8CE

The value of legs are $\left(6,6\right)$

Explanation of Solution

Looking at the graph given in the question

The given point is

  $\left(3,-3\right)\text{ and }\left(-3,3\right)$

Calculation:

The length of the legs of the right triangle can be obtained as

The point T lies on the horizontal line of $\left(3,-3\right)$ and vertical line of $\left(-3,3\right)$

Let, $\left(x_1,y_1\right)\text{ be }\left(3,-3\right)\text{ and }\left(x_2,y_2\right)\text{ be }\left(-3,3\right)$

Then,

  $\begin{array}{l}\left(x_2-x_1\right)=\left(-3-3\right)\\ \left(x_2-x_1\right)=\left(-6\right)\\ \left(x_2-x_1\right)=-6\end{array}$

The length of the legs cannot be negative

So, $\left(x_2-x_1\right)=6$

And

  $\begin{array}{l}\left(y_2-y_1\right)=\left(3-\left(-3\right)\right)\\ \left(y_2-y_1\right)=\left(3+3\right)\\ \left(y_2-y_1\right)=6\end{array}$

Hence,

The value of legs are $\left(6,6\right)$

c.

To determine

To find the length of the segment

Answer to Problem 8CE

The value of the length of the segment are $\sqrt{72}$

Explanation of Solution

Looking at the graph given in the question

The given point is

  $\left(3,-3\right)\text{ and }\left(-3,3\right)$

Calculation:

The length of the segment can be obtained as

The point T lies on the horizontal line of $\left(3,-3\right)$ and vertical line of $\left(-3,3\right)$

Let, $\left(x_1,y_1\right)\text{ be }\left(3,-3\right)\text{ and }\left(x_2,y_2\right)\text{ be }\left(-3,3\right)$

Use the distance formula between the two points to find the segment

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\left[d=\text{distance}\right]$

Put the value of $x_1=3,x_2=-3,y_1=-3\text{ and }{y}_2=3$ on the above equation

  $\begin{array}{l}d=\sqrt{{\left(-3-3\right)}^2+{\left(3-\left(-3\right)\right)}^2}\\ d=\sqrt{{\left(-6\right)}^2+{\left(3+3\right)}^2}\\ d=\sqrt{{\left(-6\right)}^2+{\left(6\right)}^2}\\ d=\sqrt{36+36}\\ d=\sqrt{72}\end{array}$

Hence,

The value of the length of the segment are $\sqrt{72}$