What is an equation of the line that passes through the point (2,-2) and is perpendicular to the line x-5y=20

To find the equation of the line that passes through the point $(2,-2)$ and is perpendicular to the line $x-5y=20$, we need to follow these steps:<br /><br />1. Find the slope of the given line $x-5y=20$.<br />2. Find the slope of the line perpendicular to the given line.<br />3. Use the point-slope form of a linear equation to find the equation of the perpendicular line.<br /><br />Step 1: Find the slope of the given line $x-5y=20$.<br /><br />To find the slope of the given line, we need to rewrite it in slope-intercept form, which is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.<br /><br />Rearranging the given equation, we get:<br />$5y = x - 20$<br />$y = \frac{1}{5}x - 4$<br /><br />The slope of the given line is $\frac{1}{5}$.<br /><br />Step 2: Find the slope of the line perpendicular to the given line.<br /><br />The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line.<br /><br />The negative reciprocal of $\frac{1}{5}$ is $-5$.<br /><br />Step 3: Use the point-slope form of a linear equation to find the equation of the perpendicular line.<br /><br />The point-slope form of a linear equation is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the slope.<br /><br />Substituting the point $(2,-2)$ and the slope $-5$ into the point-slope form, we get:<br />$y - (-2) = -5(x - 2)$<br />$y + 2 = -5(x - 2)$<br />$y + 2 = -5x + 10$<br />$y = -5x + 8$<br /><br />Therefore, the equation of the line that passes through the point $(2,-2)$ and is perpendicular to the line $x-5y=20$ is $y = -5x + 8$.