Question

find the equation of the line that is perpendicular to y = -3x + 8 and contains the point (6,5). y = \frac{1}{3}x + ?

Explanation:

Step1: Recall the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $m$ is the slope and $(x_1,y_1)$ is a point on the line. We know that the slope of the line we want to find is $m=\frac{1}{3}$ (since it is perpendicular to $y = - 3x+8$, and the slope of a perpendicular line is the negative reciprocal of $-3$, so $m=\frac{1}{3}$) and the point $(x_1,y_1)=(6,5)$.

Step2: Substitute the values into the point - slope form

Substitute $m = \frac{1}{3}$, $x_1 = 6$, and $y_1=5$ into the point - slope form:
$y - 5=\frac{1}{3}(x - 6)$

Step3: Simplify the equation

First, distribute $\frac{1}{3}$ on the right - hand side:
$y - 5=\frac{1}{3}x-2$
Then, add 5 to both sides of the equation to solve for $y$:
$y=\frac{1}{3}x-2 + 5$
$y=\frac{1}{3}x+3$

Answer:

3