First find the slope of line $k$ using its two points:

A line perpendicular to line $k$ has slope equal to the negative reciprocal of $-\frac{2}{3}$, which is $\frac{3}{2}$. Since line $m$ is perpendicular to line $k$, line $m$ has slope $\frac{3}{2}$. Any line parallel to line $m$ also has slope $\frac{3}{2}$. So the desired line through $(6,-2)$ has equation

Substitute the point $(6,-2)$:

Thus,

The equation is

which is choice B.

When a transversal cuts two parallel lines, all acute angles are congruent, all obtuse angles are congruent, and any acute angle and obtuse angle are supplementary. That means their measures add to $180$. So

Combine like terms:

Add $40$ to both sides:

Divide by $4$:

Since $x$ represents the measure of each acute angle, the equivalent expression for the measure of each acute angle is $55$.

Adjacent angles formed by two intersecting lines are supplementary, so their measures add to $180$. Let the smaller angle be $x$. Then the adjacent angle is $2x+18$. Set up the equation: $x+(2x+18)=180$. Simplify: $3x+18=180$, so $3x=162$ and $x=54$. Therefore, the smaller adjacent angle measures $54$ degrees.

When two lines intersect, adjacent angles form a linear pair, so their measures add to $180^{\circ }$. If one angle is $68^{\circ }$, then an adjacent angle is $180-68=112^{\circ }$. Therefore, the statement that must be true is that an adjacent angle has measure $112^{\circ }$.

Adjacent angles formed by two intersecting straight roads make a linear pair, so their measures add to $180$. Let the smaller angle be $x$. Then the adjacent angle is described as $2x-18$. Set up the equation:

Combine like terms:

Add $18$ to both sides:

Divide by $3$:

So, the smaller adjacent angle measures $66$ degrees.