anonymous
  • anonymous
A( \(z_A\) ) , B ( \(z_B\) ) , C ( \(z_C\) ) are vertices of right angled triangle, \(z_c\) being the orthocentre. A circle is described on AC as diameter. Find the point of intersection of the circle with hypotenuse.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
OK what can I do is just draw a diagram on the basis of the given information first... wait
anonymous
  • anonymous
@mukushla is not online this time..

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UnkleRhaukus
  • UnkleRhaukus
|dw:1352285181683:dw|
UnkleRhaukus
  • UnkleRhaukus
There it is!
anonymous
  • anonymous
If we draw a line from point C to the intersection point, the line will make 90 degrees and will be perpendicular to AB..
anonymous
  • anonymous
|dw:1352334002288:dw|
anonymous
  • anonymous
It is by the property of Semi circle..
anonymous
  • anonymous
And we are to find its coordinates.. And @UnkleRhaukus will solve this further... Ha ha ha...
anonymous
  • anonymous
Z(A), Z(B) and Z(c) are complex numbers?
UnkleRhaukus
  • UnkleRhaukus
It is by the property of Semi circle! thats right! |dw:1352285595958:dw|
anonymous
  • anonymous
Without loss of generality we can assume C is the origin ..it'll simplify the problem |dw:1352285727325:dw|
anonymous
  • anonymous
if Z(A), Z(B), Z(C) are complex numbers..
anonymous
  • anonymous
i have made some more assumption...AC=BC
anonymous
  • anonymous
otherwise use (Z-0)/(z0-0) ={ I z I / I Z0 I }e^( i pi/2)
anonymous
  • anonymous
@amistre64 and @myininaya may share their ideas ...
amistre64
  • amistre64
if i were to add anything, and im not sure how to read the notation in the problem; so i cant determine if the complex stuff is appropriate or not, but |dw:1352297044749:dw|

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