)) &A The equation of line r is y+7= -5(x+4) ndicular to line r is line s , which passes through the point (-10,-3) . What is the equation of line s? )) Write the equation in ntercept form. Write the numbers in the equation as simplified proper fractions per fractions,. or integers. square

To find the equation of line s, which is perpendicular to line r and passes through the point (-10, -3), we need to follow these steps:<br /><br />1. Find the slope of line r:<br /> The equation of line r is given as $y + 7 = -5(x + 4)$. We can rewrite this equation in slope-intercept form (y = mx + b) to find the slope of line r.<br /> $y + 7 = -5x - 20$<br /> $y = -5x - 27$<br /> The slope of line r is -5.<br /><br />2. Find the slope of line s:<br /> Since line s is perpendicular to line r, the slope of line s is the negative reciprocal of the slope of line r.<br /> The slope of line s is $\frac{1}{5}$.<br /><br />3. Use the point-slope form to find the equation of line s:<br /> The point-slope form of a line is given by $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and m is the slope.<br /> Substituting the given point (-10, -3) and the slope $\frac{1}{5}$, we get:<br /> $y - (-3) = \frac{1}{5}(x - (-10))$<br /> $y + 3 = \frac{1}{5}(x + 10)$<br /><br />4. Simplify the equation:<br /> $y + 3 = \frac{1}{5}x + 2$<br /> $y = \frac{1}{5}x - 1$<br /><br />Therefore, the equation of line s in intercept form is $y = \frac{1}{5}x - 1$.