Chapter 8.3, Problem 29WE

To determine

To graph: The line that passes through the points $K\left(-5,1\right)$ with slope $-\frac{1}{2}$ .

Explanation of Solution

Given information:

Line passes through the points $K\left(-5,1\right)$ with slope $-\frac{1}{2}$ .

Graph:

Graph the points of a line with slope $0$ ,

Points are: $\left(x_1,y_1\right)=K\left(-5,1\right)$

In general, the equation of line is: $y-y_1=m\left(x-x_1\right)$ .

Apply the values in the formula of equation of line,

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ \left(x_1,y_1\right)=\left(-5,1\right)\\ m=-\frac{1}{2}\\ y-1=-\frac{1}{2}\left(x+5\right)\end{array}$

Then, for the equation of line:

  $\begin{array}{c}y-1=-\frac{1}{2}\left(x+5\right)\\ y-1=-\frac{x}{2}-\frac{5}{2}\\ y=-\frac{x}{2}-\frac{5}{2}+1\end{array}$

Now take the constant portion:

  $\begin{array}{l}y=-\frac{x}{2}-\frac{5}{2}+1\\ y=-\frac{x}{2}-\frac{5+2}{2}\\ y=-\frac{x}{2}-\frac{3}{2}\end{array}$

Then, the equation of line is:

  $y=-\frac{x}{2}-\frac{3}{2}$

Then the graph of the line is shown below:

  

Interpretation:

Straight line equation or a linear equation, graph as straight lines, and have simple variables and expressions with no exponents on them.

If the equation with only $x$ and $y$ as opposed to say $x^2$ then dealing with a straight line equation.

In the slope intercept form of a straight line, I have $Y,M,X,B$ . They have given the value for $m$ , along with values for an $X$ and $y$ .

Then if then slope of a equation is $-\frac{1}{2}$ then the equation of the line must be: $y=-\frac{x}{2}-\frac{3}{2}$ .

By the slope of the equation and with the help of given points find the equation of line.

The graph shows that the line is straight line having the slope and the given points, that means line intersects the points $K\left(-5,1\right)$ or passes through the points.

Therefore, the equation of line $y=-\frac{x}{2}-\frac{3}{2}$ is a line passes through the points $K\left(-5,1\right)$ and have slope $-\frac{1}{2}$ .