Chapter 1, Problem 32RE

To determine

To find: an equation of the line having the given characteristics.

Answer to Problem 32RE

The equation of the line in the general form is $5x+y=11$ .

Explanation of Solution

Given:

Containing the points $\left(3,-4\right)$ and $\left(2,1\right)$ .

Calculation:

Equation of a line in slope-intercept form:

  $y=mx+b$ wherem is the slope and b is the y intercept.

Equation of a line in point-slope form

  $y-y_1=m\left(x-x_1\right)$ wherem is the slope and $x_1$ , $y_1$ is a point of the line

Equation of a line in general form

  $Ax+By=C$ where not both A and B is a positive integer. Also, though not needed, conventionally A , B and C are integers and if $A\ne 0$ , A is a positive integer.

The slope Formula:

The slope between two points $P_1\left(x_1,y_1\right)$ and $P_2\left(x_2,y_2\right)$ is:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Since $\left(3,-4\right)$ and $\left(2,1\right)$ are two points of the line, find the slope first Thus the slope of the line is

  $\begin{array}{l}m=\frac{1-\left(-4\right)}{2-3}\\ =-5\end{array}$

Equation of the line in point-slope form:

  $y-y_1=m\left(x-x_1\right)$ , where m is the slope and $\left(x_1,y_1\right)$ is a point of the line.

An equation of the line with slope -5 and containing the point (2,1) can be found by using the point-slope form with $m=-5$ , $x_1=2$ and $y_1=1$ .

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-1=-5\left(x-2\right)\text{ Substitute}\\ y-1=-5x+10\\ y=-5x+11\end{array}$

Continue further with the algebra for the general form:

  $\begin{array}{l}y=-5x+11\\ 5x+y=11\end{array}$

Conclusion:

Therefore, the equation of the line in the general form is $5x+y=11$ .