## Calculus with Applications (10th Edition)

a. $k=\frac{-1}{2}$ b. $k=\frac{-7}{2}$
Let $(x_1,y_1)=(4,-1)$ $(x_2,y_2)=(k,2)$ a. Parallel to $2x+3y=6$ By writing the equation in slope intercept form $2x+3y=6$ $y=-\frac{2}{3}x+2$ The slope is $-\frac{2}{3}$ Since the lines are parallel: $\frac{2-(-1)}{k-4}=-\frac{2}{3}$ $-2k+8=9$ $k=\frac{-1}{2}$ b. Perpendicular to $5x-2y=-1$ By writing the equation in slope intercept form $5x-2y=-1$ $y=\frac{5}{2}x+\frac{1}{2}$ The slope is $\frac{5}{2}$ Since the lines are perpendicular, $m_1.m=-1$ $\frac{5}{2}m=-1$ $m=\frac{-2}{5}$ Then, $\frac{-2}{5}=\frac{2-(-1)}{k-4}$ $-2k+8=15$ $k=\frac{-7}{2}$