Question 1165878
There's more that one way to do it.
 
CONSIDERING DISTANCES:
 
The square of the distance from a point (x,y) to the point (6,8) is 
{{{(x-6)^2+(y-8)^2}}}
In the case the point has {{{y=x}}} ,
the square of that distance to (6,8) would be
{{{(x-6)^2+(x-8)^2}}} .
The square of the distance from a point (x,y) to the point (8,0) is 
{{{(x-8)^2+(y-0)^2}}} , but if the point has {{{y=x}}} ,
the distance to (6,8) would be the square root of
{{{(x-8)^2+x^2}}} .
If the point you are looking for is at the same distance from (6.8,qnd (8,0),
those distances are the same, and their squares are the same, so
{{{(x-6)^2+(x-8)^2=(x-8)^2+x^2}}}
{{{(x-6)^2+cross((x-8)^2)=cross((x-8)^2)+x^2}}}
{{{(x-6)^2=+x^2}}}
{{{x^2-12x+36=x^2}}}
{{{cross(x^2)-12x+36=cross(x^2)}}}
{{{-12x+36=0}}}
{{{36=12x}}}
{{{36/12=x}}}
{{{highlight(x=3)}}}
The point we are looking for is {{{highlight(C(3,3))}}}
 
ANOTHER APPROACH:
The points equidistant from {{{A(6,8)}}} and {{{B(8,0)}}}
are on the bisector of segment AB.
That is the line perpendicular to AB through the midpoint of AB.
The midpoint of AB is {{{M((6+8)/2,(8+0)/2)=M(7.4)}}}
The slope of AB is {{{(8-0)/(6-8)=8/(-2)=-4}}}
The slope of a line perpendicular to AB is {{{-(1/(-4))}}}{{{"="}}}{{{1/4}}}
The equation of the line going through {{{M(7,4)}}} ,
with slope {{{M=1/4}}} is {{{y-4=(1/4)(x-7)}}} ' The intersection of that line with {{{y=x}}} is the point that we look for.
Besides {{{y=x}}} , it must comply with {{{y-4=(1/4)(x-7)}}} ,
so to find the value of {{{x}}} we solve {{{x-4=(1/4)(x-7)}}} .
{{{4(x-4)=4*(1/4)(x-7)}}}
{{{4x-16=x-7}}}
{{{4x-x=-7+16}}}
{{{3x=9}}}
{{{x=9/3}}} or {{{highlight(x=3)}}}
The point we are looking for is {{{highlight(C(3,3))}}}

{{{drawing(300,300,-1,9,-1,9,
grid(1),circle(6,8,0.1),
circle(8,0,0.1),circle(6,8,0.15),
circle(8,0,0.15),
red(circle(3,3,0.2)),
red(line(-1,-1,9,9)),
blue(line(5,12,9,-4)),
green(line(-1,2,11,5))
)}}}