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2014-10-01T05:42:32+05:30

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This is a question in co-ordinate geometry.

given straight line : L =  a x + b y + c = 0   --- equation 1
slope = m1 = - a/b

We want to find the image of a point P (x1, y1) w.r.t. the above line L. Let the image be Q(x2,y2).  Then the mid-point R of PQ lies on the line L.   Also, PQ is perpendicular to L.

So slope of PQ = m2 =  -1/m1  = b/a
          So  (y2 - y1) / b  = (x2 - x1) / a   = k (say)    -- equation 2

Mid point R of PQ = [ (x1+x2)/2, (y1+y2)/2 ],  This lies on L. so
     
           a (x1+x2)/2 + b (y1+y2)/2 + c = 0    --- equation 3

Simplifying 3 and by using k from equation 2, we get,

      k = - 2 (a x1+ b y1 + c) / (a² + b²) 
      x2 = a k + x1
       y2 = b k + y1
 
1 5 1