For Point P=(x1,y1) and Q=(x2,y2) of the co-ordinate plane, a new distance d(P,Q) is defined by d(PQ)= |x1-x2| + | y1-y2|, let O=(0,0) and A=(3,2). Prove that the set of the points in the 1st quadrant which are equidistant (with respect to the new distant) from O and A consist of the union of a line segment of finite length and an infinite ray. Sketch this set in a labelled diagram

Let the point equidistant from O and A be P(h,k)

d(O,P) = |h| + |k| = h + k (since h,k are in 1st quadrant)

d(O,A) = |h-3| + |k-2|

d(O,P) = d(O,A)

h + k = |h-3| + |k-2|

Case 1 : h>3, k>2 then |h-3| = h-3  and  |k-2| = k-2

h + k = h-3 + k-2

0 = -5  (invalid)

Case 2 : h>3, k<2 then |h-3| = h-3  and  |k-2| = 2-k

h + k = h-3 + 2-k

k = -1/2 (invalid since k>0 )

Case 3 : h<3, k>2 then |h-3| = 3-h  and  |k-2| =k-2

h + k = 3-h + k-2

h = 1/2........so locus is x = 1/2 which is a ray originating from (1/2 , 0)

Case 4 : h<3, k<2 then |h-3| = 3-h  and  |k-2| = 2-k

h + k = 3-h + 2-k

h+k = 2.5 whose locus is x+y = 2.5 which is a line segment in 1st quadrant from (0,2.5) to (2.5,0)