question_answer

The ortho centre of the \[\Delta \,OAB,\]where O is the origin,\[A(6,0)\] and \[B(3,\,3\sqrt{3})\] is

A)  \[(9/2,\,\sqrt{3}/2)\]    

B)  \[(3,\,\sqrt{3})\]

C)  \[(\,\sqrt{3},3)\]      

D)  \[(3,-\sqrt{3})\]

Correct Answer: B

Solution :

Line perpendicular to OA passing through B is \[x=3.\] Slope of \[AB=\frac{3\sqrt{3}-0}{3-6}=-\sqrt{3}\] Line perpendicular to AB through origin is \[y=\frac{1}{\sqrt{3}}x\] \[\therefore \] The point of intersection of a line \[x=3\] and \[y=\frac{1}{\sqrt{3}}x\] is \[(3,\sqrt{3}),\] which is the required orthocenter.