# For what value of k are the graphs of 12y= -3x+ 8 and 6y= kr- 5 parallel? For what value of k are they perpendicular?

Parallel at $\text{ "color(blue)"k = -3/2}$
Perpendicular at $\text{ "color(blue)"k=24}$

#### Explanation:

the given equations are $12 y = - 3 x + 8$ and $6 y = k x - 5$

Writing them first in Slope-Intercept forms

$y = - \frac{3}{12} x + \frac{8}{12} \text{ }$and $y = \frac{k}{6} x - \frac{5}{6}$

If they are parallel then slopes are equal,...therefore

$- \frac{3}{12} = \frac{k}{6}$

and $\textcolor{red}{\text{k=-3/2}}$

If they are perpendicular then the product of the slope $= - 1$

$- \frac{3}{12} \cdot \frac{k}{6} = - 1$

and $\textcolor{red}{\text{k = 24}}$

God bless....I hope the explanation is useful.