Question 336735
<pre><b>
Here is a line which is 25 units long, with all the points except Y
marked.  The point Y is somewhere beyond V but not all the way to the 
right end, since XY is less than 25 units and the line drawn is 25 units. 

{{{drawing(1000,40,-1,26,-1,1,

line(0,0,25,0), 
line(1,.2,1,-.2), line(8,.2,8,-.2),   line(15,.2,15,-.2), line(22,.2,22,-.2),
line(2,.2,2,-.2), line(9,.2,9,-.2),   line(16,.2,16,-.2), line(23,.2,23,-.2),
line(3,.2,3,-.2), line(10,.2,10,-.2), line(17,.2,17,-.2), line(24,.2,24,-.2),
line(4,.2,4,-.2), line(11,.2,11,-.2), line(18,.2,18,-.2), line(25,.2,25,-.2),
line(5,.2,5,-.2), line(12,.2,12,-.2), line(19,.2,19,-.2), 
line(6,.2,6,-.2), line(13,.2,13,-.2), line(20,.2,20,-.2), 
line(7,.2,7,-.2), line(14,.2,14,-.2), line(21,.2,21,-.2),
line(0,.2,0,-.2),
locate(0-.1,0-.2,X), locate(1-.1,0-.2,R), locate(3-.1,0-.2,S),
locate(6-.1,0-.2,T), locate(10-.1,0-.2,U), locate(13-.1,0-.2,V)
 
)}}}

I'll mark two more points on the line, A and B.  I'll put A
2 units from X and B 12 units from X

{{{drawing(1000,40,-1,26,-1,1,
line(0,.2,0,-.2),
line(0,0,25,0), 
line(1,.2,1,-.2), line(8,.2,8,-.2),   line(15,.2,15,-.2), line(22,.2,22,-.2),
line(2,.2,2,-.2), line(9,.2,9,-.2),   line(16,.2,16,-.2), line(23,.2,23,-.2),
line(3,.2,3,-.2), line(10,.2,10,-.2), line(17,.2,17,-.2), line(24,.2,24,-.2),
line(4,.2,4,-.2), line(11,.2,11,-.2), line(18,.2,18,-.2), line(25,.2,25,-.2),
line(5,.2,5,-.2), line(12,.2,12,-.2), line(19,.2,19,-.2), 
line(6,.2,6,-.2), line(13,.2,13,-.2), line(20,.2,20,-.2), 
line(7,.2,7,-.2), line(14,.2,14,-.2), line(21,.2,21,-.2),

locate(0-.1,0-.2,X), locate(1-.1,0-.2,R), locate(3-.1,0-.2,S),
locate(6-.1,0-.2,T), locate(10-.1,0-.2,U), locate(13-.1,0-.2,V),

red(locate(2-.1,1,A)), red(locate(12-.1,1,B))
 
)}}} 

The correct answer is not R because XR is 1 and XA is 2, so
R is the midpoint of XA, not XY, for Y must be beyond V.


The correct answer is not S because XS is 3 and XT is 6, so
S is the midpoint of XT, not XY, for Y must be beyond V.

The correct answer is not T because XT is 6 and XB is 12, so
T is the midpoint of XB, not XY, for Y must be beyond V.

The correct answer is not V because XV is 13 and since 2x13 = 26
V would have to be the midpoint of a line 26 units long and the
line drawn is only 25 units long.

So that only leaves point U, and it will be the midpoint of XY
if Y is 20 units from X, so now we know where Y is located, and
we can shorten the line:

{{{drawing(1000,40,-1,26,-1,1,

line(0,0,20,0), line(0,.2,0,-.2),
line(1,.2,1,-.2), line(8,.2,8,-.2),   line(15,.2,15,-.2), 
line(2,.2,2,-.2), line(9,.2,9,-.2),   line(16,.2,16,-.2), 
line(3,.2,3,-.2), line(10,.2,10,-.2), line(17,.2,17,-.2), 
line(4,.2,4,-.2), line(11,.2,11,-.2), line(18,.2,18,-.2),
line(5,.2,5,-.2), line(12,.2,12,-.2), line(19,.2,19,-.2), 
line(6,.2,6,-.2), line(13,.2,13,-.2), line(20,.2,20,-.2), 
line(7,.2,7,-.2), line(14,.2,14,-.2),

locate(0-.1,0-.2,X), locate(1-.1,0-.2,R), locate(3-.1,0-.2,S),
locate(6-.1,0-.2,T), locate(10-.1,0-.2,U), locate(13-.1,0-.2,V),
locate(20-.1,0-.2,Y),
red(locate(2-.1,1,A)), red(locate(12-.1,1,B))
 
)}}} 

The answer is V

Edwin</pre>