<PRE><B>Question 1203373</B>
&lt;br&gt;
The point (x,y) is in quadrant III, because x and y are both negative.&lt;br&gt;
Because the point is 6 units from the x-axis, y is -6.  So the point is (x,-6).&lt;br&gt;
The distance from (x,-6) to (5,7) is 15.  The difference in the y coordinates is 13, the distance is 15; find the difference in the x coordinates.&lt;br&gt;
{{{13^2+b^2=15^2}}}
{{{169+b^2=225}}}
{{{b^2=56}}}
{{{b=sqrt(56)}}}.&lt;br&gt;
The point (x,y) is {{{sqrt(56)}}} units left of x=5, so the point is (5-sqrt(56)),-6)&lt;br&gt;
The point is {{{sqrt(n)}}} units from the origin:&lt;br&gt;
{{{(5-sqrt(56))^2+6^2=(sqrt(n))^2=n}}}
{{{25+56-10sqrt(56)+36=n}}}
{{{n=117-10sqrt(56)}}} = 42.167 to 3 decimal places.&lt;br&gt;
ANSWER: (approximately) 42.167&lt;br&gt;

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