Chapter 4.3, Problem 30E

To determine

To find: The values of $a$ for which two lines are perpendicular and parallel.

Answer to Problem 30E

If linesare parallel,then $a=\frac{-2}{3}$ .

If lines are perpendicular, then $a=6$ .

Explanation of Solution

Given information:Given the equation of two lines are $6y=-2x+4\mathrm{and}2y=ax-5$ .

Calculation:Given the equation of two lines are $6y=-2x+4\mathrm{and}2y=ax-5$ .

Slope of first line is $=\frac{-2}{6}=\frac{-1}{3}$ .

Slope of second line is $=\frac{a}{2}$

If two lines are parallel then both slope are eqaual.

  $\begin{array}{l}\frac{a}{2}=\frac{-1}{3}\\ a=\frac{-2}{3}\end{array}$

If two lines are perpendicular, then product of slope is $-1$ .

  $\begin{array}{l}\frac{-1}{3}.\frac{a}{2}=-1\\ a=\frac{-3\times 2}{-1}=6\end{array}$

Therefore,if lines are parallel then $a=\frac{-2}{3}$ .

If lines are perpendicular, then $a=6$ .