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

[Vincen]

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

[Scott@TargetTestPrep]

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