Chapter 2, Problem 10CR

(a) Find the distance from $P_1=\left(-2,-3\right)\text{ to }{P}_2=\left(3,-5\right)$.

(b) What is the midpoint of the line segment from $P_1\text{ to }{P}_2$?

(c) What is the slope of the line containing the points $P_1\text{ and }{P}_2$?

To determine

To find:

a. The midpoint of a line segment joining the points $P_1$ and $P_2$.

Answer to Problem 10CR

Solution:

a. The distance from the point $P_1$ to $P_2$ is $\sqrt{41}$.

Explanation of Solution

Given:

$P_1=\left(-2, -3\right)$ and $P_2=\left(3, -5\right)$.

Formula used:

Distance formula,

$d\left(P_1,P_2\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Mid-point formula:

The midpoint of a line segment if found by averaging the $x\text{-coordinates}$ and $y\text{-coordinates}$ of the end points.

The midpoint $M$ of the line segment from $P=\left(x_1,y_1\right)$ and $Q=\left(x_2,y_2\right)$ can be found out by $M=\left(x,y\right)=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.

Slope of the line:

Equation for the slope of a line $P=\left(x_1,y_1\right)$ to $Q=\left(x_2,y_2\right)$ or from $Q$ to $P$ is $m=\frac{y_2-y_1}{x_2-x_1}=\frac{y_1-y_2}{x_1-x_2}$.

The point-slope form equation $y-y_1=m\left(x-x_1\right)$.

Calculation:

a. Distance from the point $P_1$ to $P_2$:

Substituting the given values in the distance formula, we get

$d\left(P_1,P_2\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$=\sqrt{{\left(3+2\right)}^2+{\left(-5+3\right)}^2}$

$=\sqrt{5^2+4^2}$

$=\sqrt{25+16}$

$=\sqrt{41}$

Thus, the distance from the point $P_1$ to $P_2$ is $\sqrt{41}$.

To determine

To find:

b. The midpoint of a line segment joining the points $P_1$ and $P_2$.

Answer to Problem 10CR

Solution:

b. The midpoint $M$ of the line segment is $\left(\frac{1}{2}, -4\right)$.

Explanation of Solution

Given:

$P_1=\left(-2, -3\right)$ and $P_2=\left(3, -5\right)$.

Formula used:

Distance formula,

$d\left(P_1,P_2\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Mid-point formula:

The midpoint of a line segment if found by averaging the $x\text{-coordinates}$ and $y\text{-coordinates}$ of the end points.

The midpoint $M$ of the line segment from $P=\left(x_1,y_1\right)$ and $Q=\left(x_2,y_2\right)$ can be found out by $M=\left(x,y\right)=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.

Slope of the line:

Equation for the slope of a line $P=\left(x_1,y_1\right)$ to $Q=\left(x_2,y_2\right)$ or from $Q$ to $P$ is $m=\frac{y_2-y_1}{x_2-x_1}=\frac{y_1-y_2}{x_1-x_2}$.

The point-slope form equation $y-y_1=m\left(x-x_1\right)$.

Calculation:

b. Mid-point of the line segment from $P_1$ to $P_2$:

Substituting the given values in the midpoint formula, we get

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

$M=\left(x, y\right)=\left(\frac{-2+3}{2}, \frac{-3-5}{2}\right)$

$M=\left(x, y\right)=\left(\frac{1}{2}, -4\right)$

Thus, the midpoint $M$ of the line segment is $\left(\frac{1}{2}, -4\right)$.

To determine

To find:

c. The midpoint of a line segment joining the points $P_1$ and $P_2$.

Answer to Problem 10CR

Solution:

c. The slope of the line is $-\frac{2}{5}$.

Explanation of Solution

Given:

$P_1=\left(-2, -3\right)$ and $P_2=\left(3, -5\right)$.

Formula used:

Distance formula,

$d\left(P_1,P_2\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Mid-point formula:

The midpoint of a line segment if found by averaging the $x\text{-coordinates}$ and $y\text{-coordinates}$ of the end points.

The midpoint $M$ of the line segment from $P=\left(x_1,y_1\right)$ and $Q=\left(x_2,y_2\right)$ can be found out by $M=\left(x,y\right)=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.

Slope of the line:

Equation for the slope of a line $P=\left(x_1,y_1\right)$ to $Q=\left(x_2,y_2\right)$ or from $Q$ to $P$ is $m=\frac{y_2-y_1}{x_2-x_1}=\frac{y_1-y_2}{x_1-x_2}$.

The point-slope form equation $y-y_1=m\left(x-x_1\right)$.

Calculation:

c. Slope of the line:

The slope of the line is obtained by substituting the given values in the slope formula, we get

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

$=\frac{-5-\left(-3\right)}{3-\left(-2\right)}$

$=\frac{-5+3}{3+2}$

$m=\frac{-2}{5}$

Thus, the slope of the line is $-\frac{2}{5}$.