the fixed pt is 12,0 (A)

another pt C lies on x=24

let us reduce distance CA by placing C at perpendicular to A

that is C becomes 24,0

now the other pt B lies on y=x

 

Consider the perpendicular to y=x ...which passes through A

the pt will be nothing but 6,6

now although this is shortest from A ..from C it is lon

 

Consider the perpendicular to y=x which passes through C

the pt will be nothing but 12,12

now although this is shortest to C it will be far from A

 

so the pt required is going to be midpt of 6,6 and 12,12

lying on the line y=x..because as we have seen the sum of the distance of

this line from the 2 pts A and C are going to increase on either side of this pt

and this pt as it turns out to be is

9,9