The distance #(d)# between two points #P(x_1,y_1)# and #Q(x_2,y_2)# on the xy plane is given by:
#d = sqrt((x_2-x_1)^2 +(y_2-y_1)^2)#
In this example: #d=61, P= (10,y), Q=(70,14)#
Hence: #61 = sqrt((70-10)^2+(14-y)^2)#
#61^2 = 60^2 + (14-y)^2#
#(14-y)^2 = 61^2-60^2 = (61+60)(61-60) = 121#
#y^2-28y+196 = 121#
#y^2-28y+75=0#
#(y-25)(y-3)=0#
#:. y = 25 or 3#
NB: Since the question called for a single value of #y# I suspect there was a typo. For example, if the distance between P and Q was 60 rather than 61, this would lead to a single value for y of 14, with the same logic as above as follows:
#60 = sqrt((70-10)^2+(14-y)^2)#
#60^2 = 60^2 +(14-y)^2#
#(14-y)^2=0 -> y=14#
In this case PQ is a horizontal line at #y=14#
