Chapter 1.4, Problem 4AYU

In Problems 1-4, find the following for each pair of points:

a. The distance between the points.

b. The midpoint of the line segment connecting the points.

c. The slope of the line containing the points.

d. Then interpret the slope found in part (c).

4. $(-2,-1);(3,-1)$

To determine

To find:

1. The distance between the points $\left(-2,-1\right),\left(3,-1\right)$.

2. The midpoint of the line segment joining the points.

3. The slope of the line containing the points.

4. The interpretation of the slope.

Answer to Problem 4AYU

Solution:

$\text{Distance = 5 units}$

$\text{Midpoint M }=\left(x,y\right)=(\frac{1}{2},-1)$

$\text{Slope }=\frac{0}{5};\text{ run = 5 and rise = 0}$

Explanation of Solution

Given:

The points $\left(-2,-1\right),\left(3,-1\right)$

Formula used:

Distance formula:

${\text{D(P}}_{\text{1}}{\text{, P}}_{\text{2}}\text{) }=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Midpoint $\text{M }=\left(x,y\right)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$\text{Slope }=\frac{y_2-y_1}{x_2-x_1}$

Calculation:

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

Distance $\text{d}\left(\text{AB}\right)=\sqrt{{(3-\left(-2\right))}^2+{(-1-\left(-1\right))}^2}$

$=\sqrt{{(3+2)}^2+{(-1+1)}^2}$

$=\sqrt{{(5)}^2+{(0)}^2}$

$=\sqrt{25+0}$

$=\sqrt{25}$

$\text{= 5 units}$

2. Midpoint of the line segment joining the points $\text{A}$ and $\text{B}$:

Midpoint $\text{M }=\left(x,y\right)=(\frac{3+\left(-2\right)}{2},\frac{-1+\left(-1\right)}{2})$

$=(\frac{1}{2},\frac{-2}{2})$

$=(\frac{1}{2},-1)$

3. Slope of the line containing the points $\text{A}$ and $\text{B}$ is:

$\text{Slope = }\frac{-1-\left(-1\right)}{3-\left(-2\right)}$

$=\frac{-1+1}{3+2}$

$=\frac{0}{5}$

4. The $\text{Slope }\frac{0}{5}$ means for every horizontal movement of 5 units to the right, there will be vertical movement of 0 unit. This means the line containing the points $\text{A}$ and $\text{B}$ is parallel to the $x\text{-axis}$.