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. The line given is in the slope-intercept form, which is given by the formula: $y = mx + b$, where $m$ is the slope and $b$ is the $y$-intercept. Therefore, the slope of the given line is the coefficient of $x$, so the slope is $\dfrac{1}{2}$. 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 $(1, -2)$ to plug into the formula: $y - (-2) = -\dfrac{1}{2}(x - 1)$ Simplify the equation: $y + 2 = -\dfrac{1}{2}(x - 1)$ We are asked to give the equation either in standard form or slope-intercept form. Let's rewrite this equation in slope-intercept form. First, we distribute the terms on the right side of the equation: $y + 2 = -\dfrac{1}{2}x + \dfrac{1}{2}$ Isolate $y$ by subtracting $2$ from each side of the equation: $y = -\dfrac{1}{2}x - \dfrac{3}{2}$