4) The equation of line p is y+5=-10(x+2) Line q is perpendicular to line p and passes through (-1,-1) What is the equation of line q?

To find the equation of line q, we first need to determine the slope of line p. The equation of line p is given as $y+5=-10(x+2)$. We can rewrite this in slope-intercept form (y = mx + b) to find the slope.<br /><br />$y+5=-10(x+2)$<br />$y = -10x - 20 - 5$<br />$y = -10x - 25$<br /><br />The slope of line p is -10. Since line q is perpendicular to line p, the slope of line q is the negative reciprocal of the slope of line p. Therefore, the slope of line q is $\frac{1}{10}$.<br /><br />Now, we use the point-slope form of the equation of a line, which is $y - y_1 = m(x - x_1)$, where m is the slope and $(x_1, y_1)$ is a point on the line. We are given that line q passes through the point $(-1, -1)$.<br /><br />$y - (-1) = \frac{1}{10}(x - (-1))$<br />$y + 1 = \frac{1}{10}(x + 1)$<br /><br />Therefore, the equation of line q is $y + 1 = \frac{1}{10}(x + 1)$.