Chapter 13, Problem 1CT

a.

To determine

To find: the value of $MN$ .

Answer to Problem 1CT

The value of $MN$ is $5$ .

Explanation of Solution

Given information: The points are $M\left(-2,1\right)\&N\left(2,4\right)$ .

Calculation:

  $\begin{array}{l}MN=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ MN=\sqrt{{\left(2-\left(-2\right)\right)}^2+{\left(4-1\right)}^2}\\ MN=\sqrt{{\left(4\right)}^2+{\left(3\right)}^2}\\ MN=\sqrt{16+9}=\sqrt{25}=5\end{array}$

b.

To determine

To find: the slope of $\overline{MN}$ .

Answer to Problem 1CT

The slope of $\overline{MN}$ is $\frac{3}{4}$ .

Explanation of Solution

Given information: The points are $M\left(-2,1\right)\&N\left(2,4\right)$ .

Calculation:

  $\begin{array}{l}slope=\frac{y_2-y_1}{x_2-x_1}\\ slope=\frac{4-1}{2-\left(-2\right)}=\frac{3}{4}\end{array}$

c.

To determine

To find: the midpoint of $\overline{MN}$ .

Answer to Problem 1CT

The midpoint of $\overline{MN}$ is $\left(0,\frac{5}{2}\right)$ .

Explanation of Solution

Given information: The points are $M\left(-2,1\right)\&N\left(2,4\right)$ .

Calculation:

  $\begin{array}{l}MN=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ MN=\left(\frac{-2+2}{2},\frac{4+1}{2}\right)\\ MN=\left(0,\frac{5}{2}\right)\end{array}$