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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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members; Thanked: 6 times

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

Source: Beat The GMAT — Problem Solving | Thu Jul 30, 2020 8:46 am

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8246; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

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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