Q. 10 - Precalculus Enhanced with Graphing Utilities

Question

(a) Find the distance from$P_1=(-2, -3) to P_2 = (3, -5).$

(b) What is the midpoint of the line segment from $P_1$ to $P_2$?

(c) What is the slope of the line containing the points $P_1$

and $P_2$?

Step-by-Step Solution

Answer

Part (a). The distance between point $P_{1 }$ and $P_2$ is $\sqrt{29}$.

Part (b). The midpoint of the line segment from $P_1$ to $P_2$ is $\left(\frac{1}{2},-4\right)$

Part (c). The slope of the line containing the points $P_1$ and $P_2$ is $\frac{-2}{5}$

Part (a) Step 1. Find the distance between the points

Distance formula between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$, we get:

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

So, Using the distance formula to find the distance between $P_1$ and $P_2$, we get:

$$\begin{gathered} d=\sqrt{{\left(3-(-2)\right)}^2+{\left(-5-(-3)\right)}^2} \\ d=\sqrt{5^2+{(-2)}^2} \\ d=\sqrt{25+4} \\ d=\sqrt{29} \end{gathered}$$

Part (b) Step 1. Finding the midpoint of the given line segment.

The midpoint of a line segment with their endpoints as $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as:

$$\begin{gathered} x=\frac{x_1+x_2}{2} \\ y=\frac{y_1+y_2}{2} \end{gathered}$$

So, the midpoint of the line $P_1$ and $P_2$ is given as:

$$\begin{gathered} x=\frac{-2+3}{2} \\ x=\frac{1}{2} \\ y=\frac{-3-5}{2} \\ y=\frac{-8}{2} \\ y=-4 \end{gathered}$$

So, the midpoint is $\left(\frac{1}{2},-4\right)$

Part (c) Step 1 . Finding the slope

The slope of a line containing the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as:

$m=\frac{y_2-y_1}{x_2-x_1}$

So, the slope of the line  containing $P_1$ and $P_2$ is given as:

$$\begin{gathered} m=\frac{-5-(-3)}{3-\left(-2\right)} \\ m=\frac{-5+3}{3+2} \\ m=\frac{-2}{5} \end{gathered}$$

The slope is $\frac{-2}{5}$.