text: 1. ## co-ordinate
  
  $\Delta$ABC on (Oxy) have bisector AD: x-y+2=0 , $BF\perp&space;AC$
  and BF: 4x+3y-1=0,
  $CH\perp&space;AB$ and H(-1;-1). Find co-ordinate of C ?
  
  2. Originally Posted by mp3qz
  $\Delta$ABC on (Oxy) have bisector AD: x-y+2=0 , $BF\perp&space;AC$
  and BF: 4x+3y-1=0,
  $CH\perp&space;AB$ and H(-1;-1). Find co-ordinate of C ?
  What is the meaning of this statement?
  
  3. Originally Posted by mp3qz
  $\Delta$ABC on (Oxy) have bisector AD: x-y+2=0 , $BF\perp&space;AC$
  and BF: 4x+3y-1=0,
  $CH\perp&space;AB$ and H(-1;-1). Find co-ordinate of C ?
  We are given the equation for line BF.
  We are given the equation for line AD.
  Line AC is perpendicular to BF.
  
  A little math
  $\angle CAD$ = 8deg 7min 48sec
  Which is half the angle CAB.
  
  Thus we know the equation of line AB.
  It is required that Line CH be perpendicular to AB.
  
  The equation for Line CH: y = $\frac {-3x}{4} - \, 1.75$
  
  Since H is not restricted to being on the line AB, the point C can occur anywhere on the line.
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