Line $p$ has an equation of $y = -8x + 6$ . Line $q$ , which is perpendicular to line $p$ , includes the point $(2, -2)$ . What is the equation of line $q$ $?$ $\newline$ Write the equation in slope-intercept form.

Full solution

Q. Line

p

has an equation of

y = -8x + 6

. Line

q

, which is perpendicular to line

p

, includes the point

(2, -2)

. What is the equation of line

q

?

\newline

Write the equation in slope-intercept form.

Find Slope of Line $p$ : Line $p$ has a slope of $-8$ , as we can see from its equation $y = -8x + 6$ . Since line $q$ is perpendicular to line $p$ , we need to find the slope of line $q$ , which will be the negative reciprocal of the slope of line $p$ .
Determine Slope of Line $q$ : The negative reciprocal of $-8$ is $\frac{1}{8}$ . Therefore, the slope of line $q$ is $\frac{1}{8}$ .
Calculate Y-Intercept of Line q: Now we need to find the y-intercept of line q. We know that line q passes through the point $(2, -2)$ and has a slope of $\frac{1}{8}$ . We can use the point-slope form of the equation of a line to find the y-intercept. The point-slope form is $y - y_1 = m(x - x_1)$ , where $m$ is the slope and $(x_1, y_1)$ is a point on the line.
Apply Point-Slope Form: Plugging in the slope and the point $(2, -2)$ into the point-slope form, we get $y - (-2) = \frac{1}{8}(x - 2)$ . Simplifying this, we get $y + 2 = \frac{1}{8}x - \frac{1}{4}$ .
Isolate y in Equation: To get the equation in slope-intercept form, we need to isolate y. Subtracting $2$ from both sides of the equation, we get $y = \frac{1}{8}x - \frac{1}{4} - 2$ .
Simplify Final Equation: Simplifying the equation further, we combine the constant terms $-\frac{1}{4}$ and $-2$ . Since $-2$ is the same as $-\frac{8}{4}$ , we get $y = \frac{1}{8}x - \frac{1}{4} - \frac{8}{4}$ , which simplifies to $y = \frac{1}{8}x - \frac{9}{4}$ .