Q. 8 - Geometry

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.

Step-by-Step Solution

Answer

1. The coordinate of T is (-3, -3).
2. The lengths of the legs are 6 and 6.
3. The length of the segment is $\sqrt{72}$.

Step-1 – Given

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

Step-2 – To determine

We have to find the coordinate of T.

Step-3 – Calculation

The point T lies on the horizontal line of (3, -3). So, it’s y-coordinate = -3.

The point T lies on the vertical line of (-3, 3). So, it’s x-coordinate = -3.

So, the coordinate of T is (-3, -3).

b. Step-1 – Given

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

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 (3, -3) and (-3, -3). So, the length = 3 – (-3) = 3 + 3 = 6.

The length of the vertical leg is the distance between (-3, 3) and (-3, -3). So, the length = = 3 – (-3) = 3 + 3 = 6.

Hence, the lengths of the legs are 6 and 6.

c. Step-1 – Given

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

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, 3) and (3, -3).

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

So, the length of the segment is $\sqrt{72}$.