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aksik [14]
2 years ago
7

What is the distance between parallel lines whose equations are y = 2x + 7 and y = 2x – 3?

Mathematics
1 answer:
malfutka [58]2 years ago
6 0

Answer:

The distance is \sqrt{20} or 2\sqrt{5}.

Step-by-step explanation:

Slope of parallel: 2

Slope of perpendicular: -1/2

Pick a point for y =2x + 7. For example (0,7)

Use the perpendicular slope to get to the other line to find the other point of y=2x-3. Look at the screenshot...I got (4,5).

Distance formula \sqrt{(x_{2}-x_{1} )^{2}+(y_{2} -y_{1})^2   }

Insert (0,7) and (4,5)

\sqrt{(0-4)^{2}+(7-5)^2   } = \sqrt{(-4)^{2}+(2)^2   } = \sqrt{(16)+(4)}  } = \sqrt{20}  = 2\sqrt{5}

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Step-by-step explanation:

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By definition of absolute value, you have

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or more simply,

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On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For <em>x</em> > -1, we have

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while for <em>x</em> < -1,

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More concisely,

f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x

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In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:

\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1

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