Question 566134
<pre>
The rhombus could look like any of these, even the square on the right
(a square is a rhombus).  But regardless, the diagonal CD that goes up to
the right is always a segment of the same (green) line, and the green
line is always perpendicular to the red diagonal, because the diagonals
of a rhombus are perpendicular bisectors of each other.

{{{drawing(350,350,-2,20,-2,20, graph(350,350,-2,20,-2,20),
locate(2.5,13.5,"A(6,13)"), circle(6,13,.2), green(line(-5,-5,25,25)),
locate(13.2,5.6,"B(13,6)"), circle(13,6,.2), red(line(6,13,13,6)),
line(6,13,2.5,2.5), line(13,6,2.5,2.5), line(6,13,16.5,16.5), line(13,6,16.5,16.5), locate(2.5-.5,2.5,D), locate(16.5+.3,16.5+.5,C)

 )}}}{{{drawing(350,350,-2,20,-2,20, graph(350,350,-2,20,-2,20),
locate(2.5,13.5,"A(6,13)"), circle(6,13,.2),green(line(-5,-5,25,25)),
locate(13.2,5.6,"B(13,6)"), circle(13,6,.2), red(line(6,13,13,6)),
line(6,13,7.5,7.5), line(13,6,7.5,7.5), line(6,13,11.5,11.5), line(13,6,11.5,11.5), locate(7.5-.5,7.5,D), locate(11.5+.3,11.5+.5,C)


 )}}}{{{drawing(350,350,-2,20,-2,20, graph(350,350,-2,20,-2,20),
locate(2.5,13.5,"A(6,13)"), circle(6,13,.2),green(line(-5,-5,25,25)),
locate(13.2,5.6,"B(13,6)"), circle(13,6,.2), red(line(6,13,13,6)),
line(6,13,6,6), line(13,6,6,6), line(6,13,13,13), line(13,6,13,13), locate(6-.5,6,D), locate(13+.3,13+.5,C)


 )}}}

We first find the slope of the red diagonal:

Slope formula
m = {{{(y[2]-y[1])/(x[2]-x[1])}}} with (x<sub>1</sub>, y<sub>1</sub>) = (6,13) and (x<sub>2</sub>, y<sub>2</sub>) = (13,6)

m = {{{((6)-(13))/((13)-(6))}}} = {{{(-7)/(7)}}} = -1

The slope of the green line will have the slope that is formed by
inverting {{{(-1)/1}}} as {{{1/(-1)}}} or -1 and changing the sign to +1

So the slope of the desired green line is +1.

Now we need a point that it goes through.  Since the diagonals of any
parallelogram bisect each other, we must find the midpoint of the red
diagonal.

Midpoint formula:
 
Midpoint = {{{(matrix(1,3,      (x[1]+x[2])/2,   ",", (y[1]+y[2])/2))}}} 
 
Midpoint = {{{(matrix(1,3,      ((6)+(13))/2,   ",", ((13)+(6))/2))}}}
 
Midpoint = {{{(matrix(1,3,      19/2,   ",", 19/2))}}}   

Now we use the point-slope formula:

y - y<sub>1</sub> = m(x - x<sub>1</sub>)

y - {{{19/2}}} = 1(x - {{{19/2}}})

y - {{{19/2}}} = x - {{{19/2}}}

     y = x

So that's the answer, the equation of the green line.

Edwin</pre>