Chapter 4.3, Problem 8CYU

To determine

To find: The equation of the straight line in slope-intercept form which passes through the point $\left(-1,4\right)$ and is perpendicular to the graph of line $y=3x+5$ .

Answer to Problem 8CYU

Equation of line is $y=-\frac{1}{3}x+\frac{11}{3}$ .

Explanation of Solution

Given information: The given line is $y=3x+5$ . And, another line is perpendicular to this line which passes through the point $\left(-1,4\right)$ .

Concept used: If a line is perpendicular to another line then the slope of that line will be negative reciprocal.

Calculation: Slope $\left(m\right)$ of the line perpendicular to line $y=3x+5$ , will be $\left(m=-\frac{1}{3}\right)$ .

Now, equation of line passing through the point $\left(-1,4\right)$ using point-slope form perpendicular to line $y=3x+5$ is evaluated as,

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\mathrm{......}\left(\text{Using point-slope form}\right)\\ y-4=-\frac{1}{3}\left(x+1\right)\mathrm{......}\left(\begin{array}{l}\text{Here, }m=-\frac{1}{3}\text{ and}\\ \left(x_1,y_1\right)\to \left(-1,4\right)\end{array}\right)\\ y-4=-\frac{1}{3}x-\frac{1}{3}\\ \text{or, }y=-\frac{1}{3}x+\frac{11}{3}\mathrm{......}\left(\text{Slope-intercept form}\right)\end{array}$