Find values of k if Parallel or if Perpendicular

**larry21** · June 2nd 2010, 10:47 PM

Find the values of $k$ for which the lines $kx-2y-1=0$ and $8x-ky+3=0$ are parallel.

$\frac{k}{2}=\frac{8}{k}$

${k^2}=16$

$<br /> k=\sqrt{16}=4<br />$

But my textbook says it's plus or minus 4, I only got +4 why?

B.) Are there any values of $k$ that would make the two lines perpendicular? Explain.

How would I be able to find out?

**undefined** · June 2nd 2010, 10:55 PM

Originally Posted by larry21

Find the values of $k$ for which the lines $kx-2y-1=0$ and $8x-ky+3=0$ are parallel.

$\frac{k}{2}=\frac{8}{k}$

${k^2}=16$

$<br /> k=\sqrt{16}=4<br />$

But my textbook says it's plus or minus 4, I only got +4 why?

B.) Are there any values of $k$ that would make the two lines perpendicular? Explain.

How would I be able to find out?

A)

The following is true

$x=4 \Longrightarrow x^2 = 16$

The following is not true

$x^2 = 16 \Longrightarrow x = 4$

That's because (-4)(-4) = 16.

B)

For perpendicular lines, either

1) one line is horizontal and the other is vertical, or

2) the slopes are negative reciprocals of each other.

So here we set

$\frac{k}{2}=-\frac{k}{8}$

You will get k = 0.

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