Chapter 4.4, Problem 4CYP

To determine

To find: The equation of the straight line in slope-intercept form which passes through the point $\left(4,7\right)$ and is perpendicular to the graph of line $y=\frac{2}{3}x-1$ .

Answer to Problem 4CYP

Equation of line is $y=-\frac{3}{2}x+13$ .

Explanation of Solution

Given information: The given line is $y=\frac{2}{3}x-1$ . And, another line is perpendicular to this line which passes through the point $\left(4,7\right)$ .

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 $y=\frac{2}{3}x-1$ , will be $\left(m=-\frac{3}{2}\right)$ .

Now, equation of line passing through the point $\left(4,7\right)$ using point-slope form perpendicular to line $y=\frac{1}{4}x+7$ is evaluated as,

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\mathrm{......}\left(\text{Using point-slope form}\right)\\ y-7=-\frac{3}{2}\left(x-4\right)\mathrm{......}\left(\begin{array}{l}\text{Here, }m=-\frac{3}{2}\text{ and}\\ \left(x_1,y_1\right)\to \left(4,7\right)\end{array}\right)\\ \text{or, }y-7=-\frac{3}{2}x+6\\ y=-\frac{3}{2}x+\mathrm{13......}\left(\text{Slope-intercept form}\right)\end{array}$