Question 280704
Three vertices of a parallelogram are at the points(0,0),(2,4),and(6,0).What are the coordinates of the fourth vertex?

<pre><font size = 4 color = "indigo"><b>

We plot the points:

{{{drawing(400,400,-10,10,-10,10,
 
graph(400,400,-10,10,-10,10),
locate(.2,1,"(0,0)"), locate(6.2,1,"(6,0)"), locate(2,4,"(2,4)"), 
line(0+.1,0,0-.1,0), line(0,0+.1,0,0-.1), line(0+.1,0+.1,0-.1,0-.1), line(0+.1,0-.1,0-.1,0+.1) ,
 
line(2+.1,4,2-.1,4), line(2,4+.1,2,4-.1), line(2+.1,4+.1,2-.1,4-.1), line(2+.1,4-.1,2-.1,4+.1) ,
 
line(6+.1,0,6-.1,0), line(6,0+.1,6,0-.1), line(6+.1,0+.1,6-.1,0-.1), line(6+.1,0-.1,6-.1,0+.1) )}}}
 
There are three different solutions, because we can connect those
three points in three different ways:

---------------

1. If we connect them this way:

{{{drawing(400,400,-10,10,-10,10,
 
graph(400,400,-10,10,-10,10),
locate(.2,1,"(0,0)"), locate(6.2,1,"(6,0)"), locate(2,4,"(2,4)"), 
line(0+.1,0,0-.1,0), line(0,0+.1,0,0-.1), line(0+.1,0+.1,0-.1,0-.1), line(0+.1,0-.1,0-.1,0+.1) ,
 
line(2+.1,4,2-.1,4), line(2,4+.1,2,4-.1), line(2+.1,4+.1,2-.1,4-.1), line(2+.1,4-.1,2-.1,4+.1) ,
 
line(6+.1,0,6-.1,0), line(6,0+.1,6,0-.1), line(6+.1,0+.1,6-.1,0-.1), line(6+.1,0-.1,6-.1,0+.1),

green(line(0,0,2,4)), green(line(6,0,2,4))


 )}}}

We notice that (2,4) is 2 units right of (0,0).  Therefore the 
x-coordinate of the 4th point of the parallelogram will be 2 units
left of (6,0).  That will be 6-2, or 4.

We also notice that (2,4) is 4 units above (0,0) and (6,0).  Therefore 
the y-coordinate of the 4th point of the parallelogram will be 4 units
below both (0,0,) and (6,0).  That will be 0-4, or -4.

So the 4th point is (4,-4).  The parallelogram is:
{{{drawing(400,400,-10,10,-10,10,
 
graph(400,400,-10,10,-10,10),
locate(.2,1,"(0,0)"), locate(6.2,1,"(6,0)"), locate(2,4,"(2,4)"),

line(2+.1,4,2-.1,4), line(2,4+.1,2,4-.1), line(2+.1,4+.1,2-.1,4-.1), line(2+.1,4-.1,2-.1,4+.1) ,
 
line(6+.1,0,6-.1,0), line(6,0+.1,6,0-.1), line(6+.1,0+.1,6-.1,0-.1), line(6+.1,0-.1,6-.1,0+.1),
green(line(0,0,2,4)), green(line(6,0,2,4)),
green(line(0,0,4,-4)), green(line(6,0,4,-4)),
locate(4,-4,"(4,-4)")

 )}}}

---------------

2.  If we connect them this way:

{{{drawing(400,400,-10,10,-10,10,
 
graph(400,400,-10,10,-10,10),
locate(.2,1,"(0,0)"), locate(6.2,1,"(6,0)"), locate(2,4,"(2,4)"), 
line(0+.1,0,0-.1,0), line(0,0+.1,0,0-.1), line(0+.1,0+.1,0-.1,0-.1), line(0+.1,0-.1,0-.1,0+.1) ,
 
line(2+.1,4,2-.1,4), line(2,4+.1,2,4-.1), line(2+.1,4+.1,2-.1,4-.1), line(2+.1,4-.1,2-.1,4+.1) ,
 
line(6+.1,0,6-.1,0), line(6,0+.1,6,0-.1), line(6+.1,0+.1,6-.1,0-.1), line(6+.1,0-.1,6-.1,0+.1),

green(line(0,0,2,4)), green(line(0,0,6,0))


 )}}}

We notice that (2,4) is 2 units right of (0,0).  Therefore the 
x-coordinate of the 4th point of the parallelogram will be 2 units
right of (6,0).  That will be 6+2, or 8.

We also notice that (2,4) is 4 units above (0,0) and (6,0).  Therefore 
the y-coordinate of the 4th point of the parallelogram will be 4 units
above both (0,0,) and (6,0).  That will be 0+4, or 4.

So the 4th point in this case is (4,4).  The parallelogram is:
{{{drawing(400,400,-10,10,-10,10,
 
graph(400,400,-10,10,-10,10),
locate(.2,1,"(0,0)"), locate(6.2,1,"(6,0)"), locate(2,4,"(2,4)"),

line(2+.1,4,2-.1,4), line(2,4+.1,2,4-.1), line(2+.1,4+.1,2-.1,4-.1), line(2+.1,4-.1,2-.1,4+.1) ,
 
line(6+.1,0,6-.1,0), line(6,0+.1,6,0-.1), line(6+.1,0+.1,6-.1,0-.1), line(6+.1,0-.1,6-.1,0+.1),
green(line(0,0,2,4)), green(line(0,0,6,0)),
green(line(2,4,8,4)), green(line(6,0,8,4)),
locate(8,4,"(8,4)")
 )}}}

---------------

3.  If we connect them this way:

{{{drawing(400,400,-10,10,-10,10,
 
graph(400,400,-10,10,-10,10),
locate(.2,1,"(0,0)"), locate(6.2,1,"(6,0)"), locate(2,4,"(2,4)"), 
line(0+.1,0,0-.1,0), line(0,0+.1,0,0-.1), line(0+.1,0+.1,0-.1,0-.1), line(0+.1,0-.1,0-.1,0+.1) ,
 
line(2+.1,4,2-.1,4), line(2,4+.1,2,4-.1), line(2+.1,4+.1,2-.1,4-.1), line(2+.1,4-.1,2-.1,4+.1) ,
 
line(6+.1,0,6-.1,0), line(6,0+.1,6,0-.1), line(6+.1,0+.1,6-.1,0-.1), line(6+.1,0-.1,6-.1,0+.1),

green(line(6,0,2,4)), green(line(0,0,6,0))


 )}}}

We notice that (2,4) is 4 units leftt of (6,0).  Therefore the 
x-coordinate of the 4th point of the parallelogram will be 4 units
left of (0,0).  That will be 0-4, or -4.

We also notice that (2,4) is 4 units above (0,0) and (6,0).  Therefore 
the y-coordinate of the 4th point of the parallelogram will be 4 units
above both (0,0,) and (6,0).  That will be 0+4, or 4.

So the 4th point in this case is (-4,4).  The parallelogram is:
{{{drawing(400,400,-10,10,-10,10,
 
graph(400,400,-10,10,-10,10),
locate(.2,1,"(0,0)"), locate(6.2,1,"(6,0)"), locate(2,4,"(2,4)"),

line(2+.1,4,2-.1,4), line(2,4+.1,2,4-.1), line(2+.1,4+.1,2-.1,4-.1), line(2+.1,4-.1,2-.1,4+.1) ,
 
line(6+.1,0,6-.1,0), line(6,0+.1,6,0-.1), line(6+.1,0+.1,6-.1,0-.1), line(6+.1,0-.1,6-.1,0+.1),
green(line(6,0,2,4)), green(line(0,0,6,0)),
green(line(2,4,-4,4)), green(line(0,0,-4,4)),
locate(-6,4,"(-4,4)")
 )}}}

Now you may wonder what all three solutions look like
when drawn on a graph:

{{{drawing(400,400,-10,10,-10,10,
 
graph(400,400,-10,10,-10,10),
locate(.2,1,"(0,0)"), locate(6.2,1,"(6,0)"), locate(2,4,"(2,4)"),

line(2+.1,4,2-.1,4), line(2,4+.1,2,4-.1), line(2+.1,4+.1,2-.1,4-.1), line(2+.1,4-.1,2-.1,4+.1) ,
 
line(6+.1,0,6-.1,0), line(6,0+.1,6,0-.1), line(6+.1,0+.1,6-.1,0-.1), line(6+.1,0-.1,6-.1,0+.1),
green(line(6,0,2,4)), green(line(0,0,6,0)),
green(line(2,4,-4,4)), green(line(0,0,-4,4)),
locate(-6,4,"(-4,4)"),

green(line(0,0,2,4)), green(line(0,0,6,0)),
green(line(2,4,8,4)), green(line(6,0,8,4)),
locate(8,4,"(8,4)"),

green(line(0,0,2,4)), green(line(6,0,2,4)),
green(line(0,0,4,-4)), green(line(6,0,4,-4)),
locate(4,-4,"(4,-4)")
 )}}}

They all form a big triangle, with a small triangle
inscribed insides, similar to the big triangle, whose
vertices are the midpoints of the three sides of
the big triangle.

Edwin</pre>