Question

The lines with equations $x+3y=2$ and $-2x+ky=5$ are perpendicular for $k=$

• A. $-3$
• B. $-2$
• C. $-1$
• D. $0$
• E. $2/3$

Answer 1

Answer: (E)

$2/3$

$x+3y=2,-2x+ky=5$

Slope $=\frac{-1}{3}$ Slope $=\frac{2}{k}$

Product of slopes $=-1$ (lines are perpendicular)

$=\frac{-2}{3k}=-1$

$k=\frac{2}{3}$