Question

In Exercises 4-9 state:

a. the coordinates of T.

b. the lengths of the legs of the right triangle.

c. the length of the segment shown.

Short answer

1. The coordinate of Tis (-3, -2).
2. The lengths of the legs are 2 and 4.
3. The length of the segment is $\sqrt{20}$.

Step-by-step solution

a.Step-1 – Given

The given points are (-3, 2) and (-1, -2).

Step-2 – To determine

We have to find the coordinate of T.

Step-3 – Calculation 

The point Tlies on the horizontal line of (-1, -2). So, it’s y-coordinate = -2.

The point Tlies on the vertical line of (-3, 2). So, it’s x-coordinate = -3.

So, the coordinate of Tis (-3, -2).

b.Step-1 – Given

The given points are (-3, 2) and (-1, -2).

Step-2 – To determine

We have to find the length of the legs of the right triangle.

Step-3 – Calculation 

The length of the horizontal leg is the distance between (-1, -2) and (-3, -2). So, the length = -1 – (-3) = -1 + 3 = 2.

The length of the vertical leg is the distance between (-3, 2) and (-3, -2). So, the length = = 2 – (-2) = 2 + 2 = 4.

Hence, the lengths of the legs are 2 and 4.

c.Step-1 – Given

The given points are (-3, 2) and (-1, -2).

Step-2 – To determine

We have to find the length of the shown segment.

Step-3 – Calculation 

We will use the distance formula to find the distance between (-3, 2) and (-1, -2).

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

So, the length of the segment is $\sqrt{20}$.