Question 885759
<pre>
The green line is the line 3y=-x+10

As you can see from the graphs below, there are two possible 
solutions for the rhombus.

{{{drawing(400,250,-20,20,-5,20,

line(-25,0,25,0), line(0,-25,0,25),

red(line(3,9,13.81665383,5.394448724),

line(13.81665383,5.394448724,3sqrt(13)-8,6-sqrt(13))),
locate(14.2,6,"B(?,?)"), locate(.3,1.7,"C(?,?)"),

circle(-18.8166538,9.60555128,0.15),circle(-18.8166538,9.60555128,0.13),circle(-18.8166538,9.60555128,0.11),circle(-18.8166538,9.60555128,0.09),circle(-18.8166538,9.60555128,0.07),circle(-18.8166538,9.60555128,0.05),circle(-18.8166538,9.60555128,0.03),circle(-18.8166538,9.60555128,0.01),
red(line(-8,6,3,9)),
circle(-8,6,0.15),circle(-8,6,0.13),circle(-8,6,0.11),circle(-8,6,0.09),circle(-8,6,0.07),circle(-8,6,0.05),circle(-8,6,0.03),circle(-8,6,0.01),
green(line(28,-6,-29,13)),

circle(3,9,0.15),circle(3,9,0.13),circle(3,9,0.11),circle(3,9,0.09),circle(3,9,0.07),circle(3,9,0.05),circle(3,9,0.03),circle(3,9,0.01),

circle(2.81665383,2.39444873,0.15),circle(2.81665383,2.39444873,0.13),circle(2.81665383,2.39444873,0.11),circle(2.81665383,2.39444873,0.09),circle(2.81665383,2.39444873,0.07),circle(2.81665383,2.39444873,0.05),circle(2.81665383,2.39444873,0.03),circle(2.81665383,2.39444873,0.01),

locate(3.5,10.5,"D(3,9)"), locate(-13.5,6,"A(-8,6)")

)}}}   {{{drawing(400,250,-20,20,-5,20,

line(-25,0,25,0), line(0,-25,0,25),


red(line(3,9,-7.816653826,12.60555128),

line(-7.816653826,12.60555128,-8-3sqrt(13),6+sqrt(13))),
locate(-8,14.5,"B(?,?)"), locate(-19.7,9.5,"C(?,?)"),

circle(-18.8166538,9.60555128,0.15),circle(-18.8166538,9.60555128,0.13),circle(-18.8166538,9.60555128,0.11),circle(-18.8166538,9.60555128,0.09),circle(-18.8166538,9.60555128,0.07),circle(-18.8166538,9.60555128,0.05),circle(-18.8166538,9.60555128,0.03),circle(-18.8166538,9.60555128,0.01),
red(line(-8,6,3,9)),
circle(-8,6,0.15),circle(-8,6,0.13),circle(-8,6,0.11),circle(-8,6,0.09),circle(-8,6,0.07),circle(-8,6,0.05),circle(-8,6,0.03),circle(-8,6,0.01),
green(line(28,-6,-29,13)),

circle(3,9,0.15),circle(3,9,0.13),circle(3,9,0.11),circle(3,9,0.09),circle(3,9,0.07),circle(3,9,0.05),circle(3,9,0.03),circle(3,9,0.01),

circle(2.81665383,2.39444873,0.15),circle(2.81665383,2.39444873,0.13),circle(2.81665383,2.39444873,0.11),circle(2.81665383,2.39444873,0.09),circle(2.81665383,2.39444873,0.07),circle(2.81665383,2.39444873,0.05),circle(2.81665383,2.39444873,0.03),circle(2.81665383,2.39444873,0.01),

locate(3.5,10.5,"D(3,9)"), locate(-13.5,6,"A(-8,6)")

)}}}

There is a mistake in your problem, for as you see the line BD cannot
possibly pass through the origin, yet the problem states:

"Show that BD is 3x-y=0"

But 3x-y=0 passes through the origin (0,0), so there is no way that
3x-y=0 could be the equation of BD. 

Let's find what the equation of BD really is.  

BD is parallel to AC and has the same slope.

AC has the equation  {{{3y}}}{{{""=""}}}{{{-x+10}}}

Solving for y,  {{{y}}}{{{""=""}}}{{{expr(-1/3)x+10}}}

thus the slope of AC and BD is {{{-1/3}}}

And BD passes through D(3,9)

Using the point-slope formula:

{{{y-y[1]}}}{{{""=""}}}{{{m(x-x[1])}}}

{{{y-9}}}{{{""=""}}}{{{expr(-1/3)(x-3)}}}

Multiply through by 3

{{{3y-27}}}{{{""=""}}}{{-{x-3)}}} 

{{{3y-27}}}{{{""=""}}}{{-x+3}}}

{{{3y+x}}}{{{""=""}}}{{{24}}}

This is correct equation for BD, not 3y-x=0.  

Now we must find the coordinates of the K, the intersection of the
diagonals.  Since the diagonals of a parallellogram bisect each other,
we only need to find the midpoint of one of the diagonals.  We will
need to find the coordinates of C, and then K will be the midpoint of
diagonal CD. 

Since this is a rhombus, all sides must be equal.  So we find side 
AD using the distance formula:

{{{d}}}{{{""=""}}}{{{ sqrt( (x[2]-x[1])^2 + (y[2]-y[1])^2)}}} 

{{{AD}}}{{{""=""}}}{{{ sqrt( (9-6)^2 + (3-(-8))^2)}}}
{{{AD}}}{{{""=""}}}{{{ sqrt( (3)^2 + (3+8)^2)}}}
{{{AD}}}{{{""=""}}}{{{ sqrt( 9 + (11)^2)}}}
{{{AD}}}{{{""=""}}}{{{ sqrt( 9 + 121)}}}
{{{AD}}}{{{""=""}}}{{{ sqrt(130)}}}

So the length of all 4 sides is {{{ sqrt(130)}}

Suppose the coordinates of C are C(p,q).

Since C(p,q) lies on the green line 3y = -x+10

3q = -p+10

The distance from C(p,q) to A(-8,6) must be {{{sqrt(130)}}

{{{d}}}{{{""=""}}}{{{ sqrt( (x[2]-x[1])^2 + (y[2]-y[1])^2)}}} 
{{{sqrt(130)}}}{{{""=""}}}{{{ sqrt( (p-(-8))^2 + (q-6)^2)}}}
{{{sqrt(130)}}}{{{""=""}}}{{{ sqrt( (p+8)^2 + (q-6)^2)}}} 

Square both sides:

{{{130}}}{{{""=""}}}{{{(p+8)^2 + (q-6)^2}}}
 
Now we must solve this system of equations to find p and q,
the coordinates of C

{{{system(3q = -p+10, 130=(p+8)^2 + (q-6)^2)}}}
Solve the first equation for p, p = 10-3q
Substitute in the second equation,

{{{130}}}{{{""=""}}}{{{(p+8)^2 + (q-6)^2}}}
{{{130}}}{{{""=""}}}{{{(10-3q+8)^2 + (q-6)^2}}}
{{{130}}}{{{""=""}}}{{{(18-3q)^2 + (q-6)^2}}}
{{{130}}}{{{""=""}}}{{{9q^2-108q+324 + q^2-12q+36}}}
{{{130}}}{{{""=""}}}{{{10q^2-120q+360}}}
{{{"0"}}}{{{""=""}}}{{{10q^2-120q+230}}}
Divide through by 10
{{{"0"}}}{{{""=""}}}{{{q^2-12q+23}}}
Solve by the quadratic formula and get
{{{q}}}{{{""=""}}}{{{6 +- sqrt(13)}}}

using the +, substitute in
{{{p}}}{{{""=""}}}{{{10-3q}}}
{{{p}}}{{{""=""}}}{{{ 10-3(6+sqrt(13)) ]}}}

{{{p}}}{{{""=""}}}{{{10-18-3sqrt(13))}}}
{{{p}}}{{{""=""}}}{{{-8-3sqrt(13))}}}

So point C in the graph on the right above is

{{{C(matrix(1,3,  -8-3sqrt(13), ",", 6+sqrt(13) ))}}}

using the -, substitute in
{{{p}}}{{{""=""}}}{{{10-3q}}}
{{{p}}}{{{""=""}}}{{{ 10-3(6-sqrt(13)) ]}}}

{{{p}}}{{{""=""}}}{{{10-18+3sqrt(13))}}}
{{{p}}}{{{""=""}}}{{{-8+3sqrt(13))}}}

So point C in the graph on the left above is

{{{C(matrix(1,3,  -8+3sqrt(13), ",", 6-sqrt(13) ))}}}

Now we will find the midpoint of CD which will be K,
where the two diagonals intersect.

We will find K for the graph on the left.

For the graph on the left

{{{C(matrix(1,3,  -8+3sqrt(13), ",", 6-sqrt(13) ))}}} and D(3,9)

Using the midpoint formula:

Midpoint = {{{K(matrix(1,3,(x[1]+x[2])/2, ",",(y[1]+y[2])/2))}}}

Midpoint = {{{K(matrix(1,3,(-8+3sqrt(13)+3)/2, ",",(6-sqrt(13)+9)/2))}}}

Midpoint = {{{K(matrix(1,3,(-5+3sqrt(13))/2, ",",(15-sqrt(13))/2))}}}

For the graph on the right

{{{C(matrix(1,3,  -8-3sqrt(13), ",", 6+sqrt(13) ))}}} and D(3,9)

Using the midpoint formula:

Midpoint = {{{K(matrix(1,3,(x[1]+x[2])/2, ",",(y[1]+y[2])/2))}}}

Midpoint = {{{K(matrix(1,3,(-8-3sqrt(13)+3)/2, ",",(6+sqrt(13)+9)/2))}}}

Midpoint = {{{K(matrix(1,3,(-5-3sqrt(13))/2, ",",(15+sqrt(13))/2))}}}

Edwin</pre>