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Question

The orthocentre of the OAB, where O is the origin, A6,0 and B3,33 is


A

92,32

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B

3,3

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C

3,3

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D

3,-3

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Solution

The correct option is B

3,3


Explanation for the correct option

Given that in OAB, O is the origin, A6,0 and B3,33

Let QB is the perpendicular drawn from the line OA to B

Let PO is the perpendicular drawn from the line AB to O

The intersection of the line QB and PO is the orthocenter of the triangle .

Let H be the point of orthocenter of the triangle OAB

The straight line QB is perpendicular to the x axis and meet B at 3,33

So the co-ordinate of H is 3,y

We know that two lines are perpendicular to each other is the product of their slope is equal to -1

Therefore,

SlopeofOPi.eOH×SlopeofBA=-1y-03-0×0-336-3=-1y33=1y=33y=3

So the required orthocentre =3,3

Hence, option B is the correct answer.


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