Question 550393
The diagonals of a square are perpendicular, so you need to find the equation of the line that contains AC, perpendicular to {{{x+2y=10}}}, and passing through A (1, 2).
The line that contains BD is described by the equivalent equations
{{{x+2y=10}}} --> {{{2y=-x+10}}} --> {{{y=(-1/2)x+5}}}.
So the slope of that line is {{{-1/2}}}.
The slope of all lines perpendicular to that line is
{{{(-1)/(-1/2)=2}}}.
The equation of a line with slope 2 passing through point (1, 2) is
{{{y-2=2(x-1)}}} --> {{{y-2=2x-2}}} --> {{{y=2x}}}.
So the equations of the two diagonals are
{{{y=2x}}} for the line containing AC, and
{{{x+2y=10}}} for the line containing BD.
They meet at the center of the square. The coordinates of the center of the square  can be found by solving the system formed by the to equations.
Substituting {{{y=2x}}} in {{{x+2y=10}}}
{{{x+2(2x)=10}}} -->{{{x+4x=10}}} -->{{{5x=10}}} -->{{{x=2}}}
Then, {{{y=2x=2*2=4}}}
The center of the square is (2,4). {{{drawing(300,300,-1,9,-1,9,
grid(1),
line(-1,-2,5,10),
line(-2,6,10,0),
locate(1,2,A),locate(0,5.5,D),locate(4,3.5,B),locate(2.7,6.5,C),
locate(4,8,y=2x),locate(6,2.5,x+2y=10),
blue(line(0,5,1,2)),
blue(line(1,2,4,3)),
blue(line(4,3,3,6)),
blue(line(3,6,0,5))
)}}}