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For the line whose equation is \(\dfrac{y+2-b}{x}=m+\dfrac2{x},\) \(m\) is not zero. If the line is rotated

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For the line whose equation is \(\dfrac{y+2-b}{x}=m+\dfrac2{x},\) \(m\) is not zero. If the line is rotated

by Vincen » Thu Jul 30, 2020 8:46 am

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For the line whose equation is \(\dfrac{y+2-b}{x}=m+\dfrac2{x},\) \(m\) is not zero. If the line is rotated \(90^{\circ},\) then the slope of that line would be

A. \(\dfrac1{m}\)
B. \(-\dfrac1{m}\)
C. \(m\)
D. \(-m\)
E. \(m - 2\)

[spoiler]OA=B[/spoiler]

Source: Princeton Review
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Re: For the line whose equation is \(\dfrac{y+2-b}{x}=m+\dfrac2{x},\) \(m\) is not zero. If the line is rotated

by Scott@TargetTestPrep » Wed Aug 05, 2020 10:14 am
Vincen wrote: ↑
Thu Jul 30, 2020 8:46 am
 For the line whose equation is \(\dfrac{y+2-b}{x}=m+\dfrac2{x},\) \(m\) is not zero. If the line is rotated \(90^{\circ},\) then the slope of that line would be

A. \(\dfrac1{m}\)
B. \(-\dfrac1{m}\)
C. \(m\)
D. \(-m\)
E. \(m - 2\)

[spoiler]OA=B[/spoiler]
Solution:

Multiplying the given equation by x, we have:

y + 2 - b = mx + 2

y = mx + b

Whenever a line is rotated 90 degrees, the new line is perpendicular to the original line. Since the original line has a slope of m, the new line has a slope of -1/m since -1/m is the negative reciprocal of m.

Answer: B

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