Chapter 4.3, Problem 31PPS

To determine

To Write: An equation of line in slope intercept form that passes through

$x-$ intercept of line $y=5x+3$ that is also perpendicular to required line.

Answer to Problem 31PPS

An equation of line in slope- intercept form is $y=-\frac{1}{5}x-\frac{3}{25}$

Explanation of Solution

Given information:

Line passes through $x-$ intercept and also perpendicular to line $y=5x+3$

Concept and Formula Used:

Slope intercept form $y=mx+c$ where $m$ is slope and $c$ is intercept.

Point slope form of a line: $y-y_1=m\left(x-x_1\right)$

Perpendicular lines: If lines are perpendicular then the product of their slope is $-1$

$x-$ intercept of line is point where $y$ coordinate is zero.

Calculation:

The required line is perpendicular to $y=5x+3$ .First find the slope of given line by compairing with $y=mx+c$ .

The slope of line $y=5x+3$ is $5$ .

Let the slope of required line be ‘m’. As the required line is perpendicular to $y=5x+3$

Therefore, product of slopes is $-1$

  $\begin{array}{l}m\times 5=-1\\ m=-\frac{1}{5}\end{array}$

The required line passes through $x-$ intercept of line $y=5x+3$

Putting $y=0$ in

  $\begin{array}{l}y=5x+3\\ 0=5x+3\\ -3=5x\\ x=-\frac{3}{5}\end{array}$

Therefore point from which line passes is $\left(-\frac{3}{5},0\right)$ and having slope $-\frac{1}{5}$

Using point slope form of equation of line

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ \text{where }\left(x_1,y_1\right)=\left(-\frac{3}{5},0\right)\&m=-\frac{1}{5}\\ y-\left(0\right)=-\frac{1}{5}\left(x-\left(-\frac{3}{5}\right)\right)\\ y=-\frac{1}{5}\left(x+\frac{3}{5}\right)\\ y=-\frac{1}{5}x-\frac{3}{25}\left(\text{Distributive property}\right)\end{array}$

Conclusion:

An equation of line in slope- intercept form is $y=-\frac{1}{5}x-\frac{3}{25}$