Question 1174279
<pre>
Greenstamps misread the problem.  In order for AP:AB = 2:1, P is just the
double extension of AB down to P(12,-6). P is on the infinitely long
line {{{drawing(20,40,0,3,-2,2,locate(.7,.9,AB),locate(0,2,"<->"))}}}. P could not be on the line segment (of finite length). {{{drawing(20,40,0,3,-2,2,locate(.7,.9,AB),locate(0,2.3,"___"))}}}. 
 
{{{drawing(400,2800/9,-3,15,-7,7,locate(12.35,-5.7,"P(12,-6)"),
green(line(12,0,12,-6)),
locate(.35,6.3,"A(0,6)"), locate(6.1,.8,"B(6,0)"),
locate(.2,.8,"O"),
graph(400,2800/9,-3,15,-7,7), line(0,6,12,-6), circle(12,-6,.2), 

circle(6,0,.2),circle(0,6,.2) )}}} 

I just realized there is another solution (-12,18)

{{{drawing(400,2800/9,-15,7,-7,20,locate(-12.35,18.3,"P(-12,18)"),
green(line(12,0,12,-6)),
locate(.35,6.3,"A(0,6)"), locate(6.1,.8,"B(6,0)"),
locate(.2,.8,"O"),
graph(400,2800/9,-15,7,-7,20), line(6,0,-12,18), circle(-12,18,.2), 

circle(6,0,.2),circle(0,6,.2) )}}}

Edwin</pre>