Chapter 2, Problem 16CR

To determine

To find: The slop-intercept form of equation of the line containing the points $\left(-2,4\right)$ and $\left(6,8\right)$.

Answer to Problem 16CR

Solution:

$y=\frac{1}{2}x+5$

Explanation of Solution

Given:

Line containing the points: $\left(-2,4\right)$ and $\left(6,8\right)$.

Formula used:

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:

It is given that the line containing the points are $\left(-2,4\right)$ and $\left(6,8\right)$.

Substituting the given values in the slope-intercept form of equation of a line, we get,

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

$=\frac{8-4}{6-\left(-2\right)}$

$=\frac{4}{8}$

$m=\frac{1}{2}$

Now substituting the value of slope and $x_1$ and $y_1$ in the slope-intercept form equation, we get,

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

$y-4=\frac{1}{2}x+1$

$y=\frac{1}{2}x+5$

Thus, the equation of a line is $y=\frac{1}{2}x+5$.