Q. 7 - 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 (2, -1).
2. The lengths of the legs are 6 and 2.
3. The length of the segment is $\sqrt{40}$.

a. Step-1 – Given

The given points are (-4, -1) and (2, -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 (-4, -1). So, it’s y-coordinate = -1.

The point T lies on the vertical line of (2, -3). So, it’s x-coordinate = 2.

So, the coordinate of T is (2, -1).

b. Step-1 – Given

The given points are (-4, -1) and (2, -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 (-4, -1) and (2, -1). So, the length = 2 – (-4) = 2 + 4 = 6.

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

Hence, the lengths of the legs are 6 and 2.

c. Step-1 – Given

The given points are (-4, -1) and (2, -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 (-4, -1) and (2, -3).

$\begin{array}{l}d=\sqrt{{\left(2-\left(-4\right)\right)}^2+{\left(-3-\left(-1\right)\right)}^2}\\ d=\sqrt{{\left(2+4\right)}^2+{\left(-3+1\right)}^2}\\ d=\sqrt{{\left(6\right)}^2+{\left(-2\right)}^2}\\ d=\sqrt{36+4}\\ d=\sqrt{40}\end{array}$

So, the length of the segment is $\sqrt{40}$.