Question 1023150
<pre>
From just this information.

BPCR is a rhombus, P is (0,6), C is (6,t) and B is (2,0).

we know that C(6,t) is somewhere on the vertical line whose
equation is x=6 (the green line.


{{{drawing(2400/7,400,-2,10,-4,10, graph(2400/7,400,-2,10,-4,10),
locate(-2,6.3,"P(0,6)"), locate(.1,.62,"B(2,0)"),

circle(0,6,0.15),circle(0,6,0.13),circle(0,6,0.11),circle(0,6,0.09),circle(0,6,0.07),circle(0,6,0.05),circle(0,6,0.03),circle(0,6,0.01),

circle(2,0,0.15),circle(2,0,0.13),circle(2,0,0.11),circle(2,0,0.09),circle(2,0,0.07),circle(2,0,0.05),circle(2,0,0.03),circle(2,0,0.01),

green(line(6,-5,6,11)) )}}}

we know that since it is a rhombus, PC must equal
PC, so we can draw side BP, take a compass and swing an arc like
this red one with radius BP, and find only two possibilities
for point C, th points where the arc cuts the green line:

{{{drawing(2400/7,400,-2,10,-4,10, graph(2400/7,400,-2,10,-4,10),
locate(-2,6.3,"P(0,6)"), locate(.1,.62,"B(2,0)"),

circle(0,6,0.15),circle(0,6,0.13),circle(0,6,0.11),circle(0,6,0.09),circle(0,6,0.07),circle(0,6,0.05),circle(0,6,0.03),circle(0,6,0.01),

circle(2,0,0.15),circle(2,0,0.13),circle(2,0,0.11),circle(2,0,0.09),circle(2,0,0.07),circle(2,0,0.05),circle(2,0,0.03),circle(2,0,0.01),
line(0,6,2,0), red(arc(0,6,2sqrt(40),-2sqrt(40),273,386)),

green(line(6,-5,6,11)) )}}}

So the rhombus is one of these two:

{{{drawing(2400/7,400,-2,10,-4,10, graph(2400/7,400,-2,10,-4,10),
locate(-2,6.3,"P(0,6)"), locate(.1,.62,"B(2,0)"),

locate(6.1,8.35,"C(6,t)"),

circle(0,6,0.15),circle(0,6,0.13),circle(0,6,0.11),circle(0,6,0.09),circle(0,6,0.07),circle(0,6,0.05),circle(0,6,0.03),circle(0,6,0.01),
locate(8.1,2+.35,"R(?,?)"),
circle(2,0,0.15),circle(2,0,0.13),circle(2,0,0.11),circle(2,0,0.09),circle(2,0,0.07),circle(2,0,0.05),circle(2,0,0.03),circle(2,0,0.01),
line(0,6,2,0), red(arc(0,6,2sqrt(40),-2sqrt(40),273,386)),

line(0,6,6,8),line(6,8,8,2),line(8,2,2,0),

green(line(6,-5,6,11)) )}}}{{{drawing(2400/7,400,-2,10,-4,10, graph(2400/7,400,-2,10,-4,10),locate(-2,6.3,"P(0,6)"), locate(.1,.62,"B(2,0)"),
locate(6.1,4.35,"C(6,t)"),locate(8.1,-2+.35,"R(?,?)"),
circle(0,6,0.15),circle(0,6,0.13),circle(0,6,0.11),circle(0,6,0.09),circle(0,6,0.07),circle(0,6,0.05),circle(0,6,0.03),circle(0,6,0.01),

circle(2,0,0.15),circle(2,0,0.13),circle(2,0,0.11),circle(2,0,0.09),circle(2,0,0.07),circle(2,0,0.05),circle(2,0,0.03),circle(2,0,0.01),

line(0,6,2,0), red(arc(0,6,2sqrt(40),-2sqrt(40),273,386)),
line(0,6,6,4),line(6,4,8,-2),line(8,-2,2,0),

green(line(6,-5,6,11)) )}}}

Since we are told that PR = 2BC, we know it must be the second one,
not the first.

{{{drawing(2400/7,400,-2,10,-4,10, graph(2400/7,400,-2,10,-4,10),locate(-2,6.3,"P(0,6)"), locate(.1,.62,"B(2,0)"),
locate(6.1,4.35,"C(6,t)"),locate(8.1,-2+.35,"R(?,?)"),
circle(0,6,0.15),circle(0,6,0.13),circle(0,6,0.11),circle(0,6,0.09),circle(0,6,0.07),circle(0,6,0.05),circle(0,6,0.03),circle(0,6,0.01),

circle(2,0,0.15),circle(2,0,0.13),circle(2,0,0.11),circle(2,0,0.09),circle(2,0,0.07),circle(2,0,0.05),circle(2,0,0.03),circle(2,0,0.01),

line(0,6,2,0), red(arc(0,6,2sqrt(40),-2sqrt(40),273,386)),
line(0,6,6,4),line(6,4,8,-2),line(8,-2,2,0),

green(line(6,-5,6,11)) )}}}

Since all the sides of a rhombus are equal in
length, we use the distance formula to find BP:

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

{{{BP=sqrt(((0)-(2))^2+((6)-(0))^2)=sqrt(4^2+6^2)=sqrt(16+36)=sqrt(40)}}}

We use the distance formula again to set the
BP = PC =  &#8730;40

{{{sqrt(40)=sqrt((t-6)^2+(6-0)^2)}}}

{{{sqrt(40)=sqrt((t-6)^2+36)}}}

Square both sides:

{{{40=(t-6)^2+36}}}

{{{4=(t-6)^2}}}

take square roots of both sides:

{{{"" +- 2=t-6}}}

{{{6 +- 2=t}}}

So t=4 or 8.  Obviously the answer we 
want is  C(6,4),

So t=4

We can get the coordinates of R just by
counting units.  We know that B is 6
units down and 2 units right from P.
So R must be 6 units down and 2 units
right from C.  That would give us R(8,-2).

We could also get the coordinates of R 
this way.  We know that C is 6
units right and 2 units down from P.
So R must be 6 units right and 2 units
down from B.  That would also give us 
R(8,-2).

If we like can check to make sure that 

PR = 2BC and PR =8sqrt 2

by using the distance formula, but
that's obvious if this is a legitimate
problem.

To find the equation of PR, the blue line below:

{{{drawing(2400/7,400,-2,10,-4,10, graph(2400/7,400,-2,10,-4,10),locate(-2,6.3,"P(0,6)"), locate(.1,.62,"B(2,0)"),
locate(6.1,4.35,"C(6,t)"),locate(8.1,-2+.35,"R(8,-2)"),
circle(0,6,0.15),circle(0,6,0.13),circle(0,6,0.11),circle(0,6,0.09),circle(0,6,0.07),circle(0,6,0.05),circle(0,6,0.03),circle(0,6,0.01),

circle(2,0,0.15),circle(2,0,0.13),circle(2,0,0.11),circle(2,0,0.09),circle(2,0,0.07),circle(2,0,0.05),circle(2,0,0.03),circle(2,0,0.01),

line(0,6,2,0), red(arc(0,6,2sqrt(40),-2sqrt(40),273,386)),
line(0,6,6,4),line(6,4,8,-2),line(8,-2,2,0),
blue(line(-3,9,16,-10.03)),
green(line(6,-5,6,11)) )}}}

To find the equation of the PR, the blue line, we use 
the slope formula to find the slope:

Slope formula:
m = {{{(y[2]-y[1])/(x[2]-x[1])}}}
where 

P(x<sub>1</sub>,y<sub>1</sub>) = P(0,6)
and where R(x<sub>2</sub>,y<sub>2</sub>) = R(8,-2)

m = {{{((-2)-(6))/((8)-(0))}}}
m = {{{(-8)/(-8)}}}
m = 1

Then we use the point-slope formula:
y - y<sub>1</sub> = m(x - x<sub>1</sub>)
where (x<sub>1</sub>,y<sub>1</sub>) = (0,6)

y - 6 = 1(x - 0)
y - 6 = x
    y = x + 6

Edwin</pre>