Question 250812
<pre><font size = 4 color = "indigo"><b>
First draw the three points and connect them, like this

{{{drawing(400,400,-2,9,-2,9,

graph(400,400,-2,9,-2,9), line(2,1,1,4), line(2,1,7,2),
locate(2-.3,1,"A(2,1)"), locate(1-.7,4+.6,"B(1,4)"), locate(7+.2,2+.2,"C(7,2)")

 )}}}

Sketch in the rest of the parallelogram, 
a green line from B to about where you think D
should be, and green line from C to that point D.

{{{drawing(400,400,-2,9,-2,9,

graph(400,400,-2,9,-2,9), line(2,1,1,4), line(2,1,7,2),
locate(2-.3,1,"A(2,1)"), locate(1-.7,4+.6,"B(1,4)"), locate(7+.2,2+.2,"C(7,2)"),
green(line(7,2,6,5)), green(line(6,5,1,4)), green(locate(6+.2,5+.4,D)) 

 )}}}

Think about how to to get from A to B going 
left and then up.  You'd start at A and go left
1 unit horizontally until you get to a point 
exactly under B, like the short blue line below:

{{{drawing(400,400,-2,9,-2,9,

graph(400,400,-2,9,-2,9), line(2,1,1,4), line(2,1,7,2),
locate(2-.3,1,"A(2,1)"), locate(1-.7,4+.6,"B(1,4)"), locate(7+.2,2+.2,"C(7,2)"),
green(line(7,2,6,5)), green(line(6,5,1,4)), green(locate(6+.2,5+.4,D)),
blue(line(2,1,1,1)) 

 )}}}

Notice that when you get to the left end of that blue
line you are now directly over the 1 on the x-axis,
and directly to the right the 1 on the y-axis.

Then from there you'd have to turn and go 3 units up 
vertically to point B, like this other vertical blue line:

{{{drawing(400,400,-2,9,-2,9,

graph(400,400,-2,9,-2,9), line(2,1,1,4), line(2,1,7,2),
locate(2-.3,1,"A(2,1)"), locate(1-.7,4+.6,"B(1,4)"), locate(7+.2,2+.2,"C(7,2)"),
green(line(7,2,6,5)), green(line(6,5,1,4)), green(locate(6+.2,5+.4,D)),
blue(line(2,1,1,1)), blue(line(1,1,1,4)) 

 )}}}

Notice that you are directly right of the the 4 on the y-axis.

Next do the same thing going from C to D. You had to go 1 unit
left in going from A to get under B, so you'll have to also go
1 unit to the left of C to get right under D, like this:

{{{drawing(400,400,-2,9,-2,9,

graph(400,400,-2,9,-2,9), line(2,1,1,4), line(2,1,7,2),
locate(2-.3,1,"A(2,1)"), locate(1-.7,4+.6,"B(1,4)"), locate(7+.2,2+.2,"C(7,2)"),
green(line(7,2,6,5)), green(line(6,5,1,4)), green(locate(6+.2,5+.4,D)),
blue(line(2,1,1,1)), blue(line(1,1,1,4)), blue(line(7,2,6,2)) 

 )}}}

Notice that you are exactly above the 6 on the x-axis, and
exatly to the right of 2 on the y-axis. Now since you had to
go up 3 units to get from A to B, you'll need to go 3 units 
up from there to get to D, like this:

{{{drawing(400,400,-2,9,-2,9,

graph(400,400,-2,9,-2,9), line(2,1,1,4), line(2,1,7,2),
locate(2-.3,1,"A(2,1)"), locate(1-.7,4+.6,"B(1,4)"), locate(7+.2,2+.2,"C(7,2)"),
green(line(7,2,6,5)), green(line(6,5,1,4)), green(locate(6+.2,5+.4,D)),
blue(line(2,1,1,1)), blue(line(1,1,1,4)), blue(line(7,2,6,2)),
blue(line(6,2,6,5)) 

 )}}} 

Since you had to go 3 up from being beside 2 on the y-axis, you
end up being exactly beside the 5 on the y-axis. So the point D
must be (6,5).

{{{drawing(400,400,-2,9,-2,9,

graph(400,400,-2,9,-2,9), line(2,1,1,4), line(2,1,7,2),
locate(2-.3,1,"A(2,1)"), locate(1-.7,4+.6,"B(1,4)"), locate(7+.2,2+.2,"C(7,2)"),
green(line(7,2,6,5)), green(line(6,5,1,4)), green(locate(6+.2,5+.4,"D(6,5)")),
blue(line(2,1,1,1)), blue(line(1,1,1,4)), blue(line(7,2,6,2)),
blue(line(6,2,6,5)) 

 )}}}

So now you see you could have done it in your head. You would
just notice that the point B(1,4) is 1 unit left and 3 units up
from point A(2,1), so the point D has to be 1 unit left
and 3 units up from point C(7,2), so you'd subtract 1 from the
x-coordinate of C, getting 6, and then add 3 to the y-coordinate 
of C to get 5, and you get D(6,5). Pretty easy, huh?

Edwin</pre>