## Algebra 1: Common Core (15th Edition)

$y = \frac{1}{3}x + 4$
For perpendicular lines, the product of the slopes of the two lines is $-1$. The equation of the line that we are given is in slope-intercept form, which is given by the formula: $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. The slope of the line given is $-3$. The product of the slopes of the two lines must equal $-1$, so let us set up the equation to find the slope $m$ of the unknown line: $-3(m) = -1$ Divide each side by $-3$ to solve for $m$: $m = \frac{-1}{-3}$ Divide both the numerator and denominator by $-1$ to simplify the fraction: $m = \frac{1}{3}$ Now that we have the slope of the unknown line, we can plug this and the point that we are given $(3, 5)$ into point-slope form, which is given by the formula: $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is a point on that graph. Let us plug in the values: $y - 5 = \frac{1}{3}(x - 3)$ Let's convert this equation into point-intercept form: Use the distributive property: $y - 5 = \frac{1}{3}x + \frac{1}{3}(-3)$ Multiply to simplify: $y - 5 = \frac{1}{3}x - \frac{3}{3}$ Simplify the fraction by dividing the numerator by the denominator: $y - 5 = \frac{1}{3}x - 1$ To isolate $y$, add $5$ to each side of the equation: $y = \frac{1}{3}x - 1 + 5$ Add to simplify: $y = \frac{1}{3}x + 4$ This is the equation of the line that we are looking for in slope-intercept form.