Question

The equation of line n is 8x - 14y + 3 = 0. What is k if the graphs of n and kx - 7y + 10 = 0 are perpendicular?

Answer 1

First put the given equations into the form, y = mx + c, as follows:
$y=\frac{4}{7}x+\frac{3}{14}$
$y=\frac{k}{7}x+\frac{10}{7}$
For perpendicularity, the product of the slopes must equal negative one.
$\frac{k}{7}\times\frac{4}{7}=-1$
Solving for k, we get:
$k=-\frac{49}{4}$
The correct choice is C.