WE HAVE 2 LINES AB AND BC WHERE IS AB PERPENDICULAR TO X AXIS AND BC IS PARALLEL TO X AXIS(LIKE RECTANGLE)THEN product of SLOPE OF two PERPEND. lineS?LIKE THIS CASE IS POSSIBLE IN CASE OF A...

WE HAVE 2 LINES AB AND BC WHERE IS AB PERPENDICULAR TO X AXIS AND BC IS PARALLEL TO X AXIS(LIKE RECTANGLE)THEN product of SLOPE OF two PERPEND. lineS?

LIKE THIS CASE IS POSSIBLE IN CASE OF A RECTANGLE CONSIDERING OABC(HERE O IS THE ORIGIN , N OABC IS A RECTANGLE) SO HERE THE OF SLOPE OF AB WILL BE INFINITY (SINCE IT IS PERPENDICULAR TO X AXIS ) ALSO SLOPE OF BC WILL BE 0 (SINCE IT IS MAKING 0 DEGREE ANGLE WITH X AXIS ) . SO THE PRODUCT OF INFINITY X 0 IS NOT -1 .

Asked on by megha5

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quantatanu | Student, Undergraduate | (Level 1) Valedictorian

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Its a brilliant question. In such situation, where you encounter Infinities, you must always choose Limit. As for example in this case, suppose the rectangle is slightly rotated by an angle "theta" where

"theta ---> 0"

So now line BC makes an angle "theta" with x-axis and 

           line AB makes an angle "Pi/2 + theta" with x-axis

 


So slope of BC (m_bc) =  tan(theta)


   slope of AB (m_ab) = tan(Pi/2 + theta)

 

Hence,

m_bc * m_ab = tan(theta) * tan(Pi/2 + theta)

         = tan(theta) * [sin(Pi/2 +theta]/[cos(Pi/2 + theta]

          = tan(theta) * [Cos(theta)]/[-Sin(theta)]

          = tan(theta) * [-1/tan(theta)]

          = -1

Now take the limit "theta----> 0"

so

Limit (theta-->0) [m_bc * m_ab] = Limit(theta-->0)[-1]

                                                      = -1

 

Now you can see this limiting procedure solves the puzzel.

 

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