Chapter 1, Problem 30RE

To determine

To find: an equation of the line

Answer to Problem 30RE

Theequations is $y=-\frac{5x}{2}+5$ and the general form is $5x+2y=10$ .

Explanation of Solution

Given:

  $x-intercept=2; containingthepoint\left(4,-5\right)$

Calculation:

Equation of a line in slope-intercept form:

  $y=mx+b\text{ where }m\text{ is the slope and }b\text{ is the }y\text{ intercept}\text{.}$

Equation of a line in point-slope form

  $y-y_1=m\left(x-x_1\right)\text{ where }m\text{ is the slope and }{x}_1,y_1\text{ is a point of the line}$

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 the x -intercept is 2, (2,0) is a point of the line. It is also given that (4,-5) is a point of the line, find the slope first.

Thus the slope of the line is

  $\begin{array}{l}m=\frac{-5-0}{4-2}\\ =-\frac{5}{2}\end{array}$

Equation of the line in point-slope form:

  $y-y_1=m\left(x-x_1\right)\text{ where }m\text{ is the slope and }{x}_1,y_1\text{ is a point of the line}$ .

An equation of the line with slope $-\frac{5}{2}$ and containing the point (2,0) can be found by using the point-slope form with $m=-\frac{5}{2}$ , $x_1=2$ and $y_1=0$ .

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-0=-\frac{5}{2}\left(x-2\right)\text{ Substitute}\\ y=-\frac{5x}{2}+5\end{array}$

Continue further with the algebra for the general form:

  $\begin{array}{l}y=-\frac{5x}{2}+5\\ 2y=-5x+5\\ 5x+2y=10\end{array}$

Conclusion:

Therefore, the equations is $y=-\frac{5x}{2}+5$ and the general form is $5x+2y=10$ .