Chapter 13.1, Problem 9CE

a.

To determine

To find the coordinates of T

Answer to Problem 9CE

The value of T is $\left(-3,-2\right)$

Explanation of Solution

Looking at the graph given in the question

The given point is

  $\left(-1,-2\right)\text{ and }\left(-3,2\right)$

Calculation:

The coordinates of T can be obtained as

The point T lies on the horizontal line of $\left(-1,-2\right)$ and vertical line of $\left(-3,2\right)$

Let, $\left(x_1,y_1\right)\text{ be }\left(-1,-2\right)\text{ and }\left(x_2,y_2\right)\text{ be }\left(-3,2\right)$

Then the point T is $\left(x_2,y_1\right)=\left(-3,-2\right)$

Hence,

The value of T is $\left(-3,-2\right)$

b.

To determine

To find the length of the legs of the right triangle

Answer to Problem 9CE

The value of legs are $\left(2,4\right)$

Explanation of Solution

Looking at the graph given in the question

The given point is

  $\left(-1,-2\right)\text{ and }\left(-3,2\right)$

Calculation:

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

The point T lies on the horizontal line of $\left(-1,-2\right)$ and vertical line of $\left(-3,2\right)$

Let, $\left(x_1,y_1\right)\text{ be }\left(-1,-2\right)\text{ and }\left(x_2,y_2\right)\text{ be }\left(-3,2\right)$

Then,

  $\begin{array}{l}\left(x_2-x_1\right)=\left(-3-\left(-1\right)\right)\\ \left(x_2-x_1\right)=\left(-3+1\right)\\ \left(x_2-x_1\right)=-2\end{array}$

The length of the legs cannot be negative

So, $\left(x_2-x_1\right)=2$

And

  $\begin{array}{l}\left(y_2-y_1\right)=\left(2-\left(-2\right)\right)\\ \left(y_2-y_1\right)=\left(2+2\right)\\ \left(y_2-y_1\right)=4\end{array}$

Hence,

The value of legs are $\left(2,4\right)$

c.

To determine

To find the length of the segment

Answer to Problem 9CE

The value of the length of the segment are $\sqrt{20}$

Explanation of Solution

Looking at the graph given in the question

The given point is

  $\left(-1,-2\right)\text{ and }\left(-3,2\right)$

Calculation:

The length of the segment can be obtained as

The point T lies on the horizontal line of $\left(-1,-2\right)$ and vertical line of $\left(-3,2\right)$

Let, $\left(x_1,y_1\right)\text{ be }\left(-1,-2\right)\text{ and }\left(x_2,y_2\right)\text{ be }\left(-3,2\right)$

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 $x_1=-1,x_2=-3,y_1=-2\text{ and }{y}_2=2$ on the above equation

  $\begin{array}{l}d=\sqrt{{\left(-3-\left(-1\right)\right)}^2+{\left(2-\left(-2\right)\right)}^2}\\ d=\sqrt{{\left(-3+1\right)}^2+{\left(2+2\right)}^2}\\ d=\sqrt{{\left(-2\right)}^2+{\left(4\right)}^2}\\ d=\sqrt{4+16}\\ d=\sqrt{20}\end{array}$

Hence,

The value of the length of the segment are $\sqrt{20}$