1
IIT-JEE 2003 Screening
+3
-0.75
Orthocentre of triangle with vertices $$\left( {0,0} \right),\left( {3,4} \right)$$ and $$\left( {4,0} \right)$$ is
A
$$\,\,\left( {3,{5 \over 4}} \right)$$
B
$$\left( {3,12} \right)$$
C
$$\left( {3,{3 \over 4}} \right)$$
D
$$\left( {3,9} \right)$$
2
IIT-JEE 2002 Screening
+3
-0.75
Let $$0 < \alpha < {\pi \over 2}$$ be fixed angle. If $$P = \left( {\cos \theta ,\,\sin \theta } \right)$$ and $$Q = \left( {\cos \left( {\alpha - \theta } \right),\,\sin \left( {\alpha - \theta } \right)} \right),$$ then $$Q$$ is obtained from $$P$$ by
A
clockwise rotation around origin through an angle $$\alpha$$
B
anticlockwise rotation around origin through an angle $$\alpha$$
C
reflection in the line through origin with slope tan $$\alpha$$
D
reflection in the line through origin with slope tan $$\left( {\alpha /2} \right)$$
3
IIT-JEE 2002 Screening
+3
-0.75
Let $$P = \left( { - 1,\,0} \right),\,Q = \left( {0,\,0} \right)$$ and $$R = \left( {3,\,3\sqrt 3 } \right)$$ be three points.
Then the equation of the bisector of the angle $$PQR$$ is
A
$${{\sqrt 3 } \over 2}x + y = 0$$
B
$$x + \sqrt 3 y = 0$$
C
$$\sqrt 3 x + y = 0$$
D
$$x + {{\sqrt 3 } \over 2}y = 0$$
4
IIT-JEE 2002 Screening
+3
-0.75
A straight line through the origin $$O$$ meets the parallel lines $$4x+2y=9$$ and $$2x+y+6=0$$ at points $$P$$ and $$Q$$ respectively. Then the point $$O$$ divides the segemnt $$PQ$$ in the ratio
A
$$1 : 2$$
B
$$3 : 4$$
C
$$2 : 1$$
D
$$4 : 3$$
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