[MOVED] perpendicular line...what is k?

The equation of the line m is 8x - 14y + 3 = 0

a) For what value of k is the graph of kx - 7y + 10 = 0 parallel to line m?

My work:

. . .slope:
. . . . .-A/B = -8/-12 = 8/12 = 4/7
. . . . .-A/B = -?/-7
. . . . .k = 4

b) What is k if the graphs of m and kx - 7y + 10 = 0 are perpendicular?

I don't know how to do this part.

The answer to part b is -49/4, but I have no idea how to get the answer. I thought all you had to do was find the opposite reciprocal but it turns out to be -7/4 NOT -49/4

Please help! Thank you!

Solve m for y:

\(\displaystyle \L\\y=\frac{4x}{7}+\frac{3}{14}\)

The other equation is:

\(\displaystyle \L\\y=\frac{kx}{7}+\frac{10}{7}\)

Since they are parallel, the slopes are the same. in the latter equation, what value of k makes the slope 4/7? .

If they are perpendicular, the slope of the k equation is the negative reciprocal of the m equation.

\(\displaystyle \L\\\frac{kx}{7}+\frac{10}{7}\)

What value of k makes the slope -7/4?.

\(\displaystyle \frac{k}{7}=\frac{-7}{4}\)

Solve for k.

Here is another way to look at it:
The line \(\displaystyle Ax + By + C = 0\) has slope \(\displaystyle \frac{{ - A}}{B}\) if \(\displaystyle AB \not= 0\).
Lines with the same slope are parallel.
Lines who’s slopes multiple to \(\displaystyle - 1\) are perpendicular.
So line \(\displaystyle Ax + By + C = 0\) is perpendicular to line \(\displaystyle Bx -Ay + C = 0.\)