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 $P_2$?

Short 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$\frac{-2}{5}$

Step-by-step solution

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:

role="math" localid="1645947305764" $$\begin{gathered} x=\frac{-2+3}{2} \\ x=\frac{1}{2} \\ y=\frac{-3-5}{2} \\ y=\frac{-8}{2} \\ y=-4 \end{gathered}$$

So, the midpoint is role="math" localid="1645947334566" $\left(\frac{1}{2},-4\right)$

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$\frac{-2}{5}$.