Are the lines $y = 8 x - 5, y = \frac{1}{8} x + 1$ $y=8x-5, y=1/8x+1$ and $8 x + y = 2$ $8x+y=2$ parallel or perpendicular?

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Are the lines $y = 8 x - 5, y = \frac{1}{8} x + 1$ $y=8x-5, y=1/8x+1$ and $8 x + y = 2$ $8x+y=2$ parallel or perpendicular?

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Answer 1 · 2018-06-06T04:01:57

The second and third lines are perpendicular

Explanation:

If you restructure the third equation by subtracting $8 x$ $8x$ from both sides, you'll get

$y = - 8 x + 2$ $y=-8x+2$

With

$y = m x + b$ $y=mx+b" "$ [here $m =$ $m =$ slope]

you know the slope of all three lines: $8$ $8$ , $\frac{1}{8}$ $1/8$ , and $- 8$ $-8$ , respectively.

Two lines are perpendicular when their slopes are negative reciprocals . You can see that two slopes are negative reciprocals if $- 1$ $-1$ divided by one slope equals the other slope.

Because $- 1$ $-1$ divided by $- 8$ $-8$ equals $\frac{1}{8}$ $1/8$ , you know that the second and third equations have negative reciprocal slopes, therefore they are perpendicular.

The first equation isn't perpendicular to either of the other two lines because $- 1$ $-1$ divided by $8$ $8$ is $- \frac{1}{8}$ $-1/8$ , which does not match any of the other slopes. It isn't parallel either (parallel lines have the same slope).

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Are the lines $y = 8 x - 5, y = \frac{1}{8} x + 1$ $y=8x-5, y=1/8x+1$ and $8 x + y = 2$ $8x+y=2$ parallel or perpendicular?

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Explanation:

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Explanation:

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