## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$y = \dfrac{1}{2}x + \dfrac{3}{2}$
With lines that are perpendicular to each other, the product of their slopes is $-1$; one slope is the negative reciprocal of the other. If we want to find the slope of a line that is perpendicular to a given line, we must first find the slope of the given line. This given line is written in standard form, so we want to rewrite this line in slope-intercept form, which is given by the following formula: $y = mx + b$, where $m$ is the slope of the line and $b$ is the $y$-intercept. We can rewrite our equation by isolating our $y$ term. First, we subtract $2x$ from each side of the equation: $y = -2x + 2$ Therefore, the slope of the given line is the coefficient of $x$, so the slope is $-2$. Let us set up an equation to find the slope of the line that is perpendicular to the given line by multiplying the two slopes to yield $-1$. Let $m$ be the slope of the perpendicular line: $(-2)(m) = -1$ Divide both sides by $-2$: $m = \dfrac{1}{2}$ Let us plug this slope and the point we are given into the point-slope form of the equation, which is given by the formula: $y - y_1 = m(x - x_1)$, where $m$ is the slope of the line and $(x_1, y_1)$ is a point on that line. Let us use the point $(-3, 0)$ to plug into the formula: $y - 0 = \frac{1}{2}(x - (-3))$ Simplify the equation: $y = \dfrac{1}{2}(x + 3)$ We are asked to give the equation either in standard form or slope-intercept form. Let us rewrite this equation in slope-intercept form. Distribute the terms on the right side of the equation: $y = \dfrac{1}{2}x + \dfrac{3}{2}$