Question 397657
<pre><font face = "batangche" color = "indigo" size = 4><b>

The other tutor is right! Here's a bit more geometrical drawings:  

Suppose this is the line segment.

0------------x----------y------------a

Then the sides of the triangle are x, y-x, and a-y

The sum of any two sides must be greater than the third side.

So

x + (y-x) > a-y,    x + (a-y) > y-x,     (y-x) + (a-y) > x      y > x
x + y - x > a-y     x + a - y > y-x            y-x+a-y > x 
        y > a-y       2x - 2y > -a                -x+a > x   
       2y > a        -2x + 2y < a                    a > 2x
        y > a/2       2y - 2x < a                  a/2 > x   
                       2(y-x) < a                    x < a/2   
                          y-x < {{{a/2}}}          

We also know that y < a  and x > 0

So we have a system of inequalities:

{{{system(y<a,y>a/2,y-x<a/2,x>0, x<a/2, y > x)}}}

We draw the boundary lines:


{{{drawing(400,400,-.3,.9,-.3,.9, graph(400,400,-.3,.9,-.3, .9), 

line(.4,.02,.4,-.02), line(.8,.02,.8,-.02), locate(.39,-.02,a/2),
locate(.79,-.02,a), line(.02,.4,-.02,.4),line(.02,.8,-.02,.8),

locate(-.05,.46,a/2), locate(-.05,.83,a), line(0,0,2,2),
locate(.15,.5,SOLUTION),
line(0.01,.41,.39,.41),line(.39,.79,.39,.41),line(0.015,.41,.4,.785),

green(line(0,.4,2,2.4),line(0,.8,2,.8),line(0,.4,2,.4),line(.4,0,.4,.8))
 



)}}}

and find that the solution to the system of inequalities is the triangle
marked "SOLUTION".  It's area is 

{{{A=expr(1/2)*base*height=expr(1/2)*expr(a/2)*expr(a/2)}}}{{{""=""}}}{{{a^2/8}}}

So the numerator of the probability is the area of that triangle, for
the coordinates of any point within that triangle will represent points
on the original line:


0------------x----------y------------a

that satisy the system of inequalities.

The denominator of the probability is the area of this triangle:

{{{drawing(400,400,-.3,.9,-.3,.9, graph(400,400,-.3,.9,-.3, .9), 

line(.4,.02,.4,-.02), line(.8,.02,.8,-.02), locate(.39,-.02,a/2),
locate(.79,-.02,a), line(.02,.4,-.02,.4),line(.02,.8,-.02,.8),

locate(-.05,.46,a/2), locate(-.05,.83,a),

green(triangle(0,0,.8,.8,0,.8))
 



)}}}
 
Because the coordinates of any point inside this triangle represents 
a pair of values for x and y that could have been picked on this line:

0------------x----------y------------a
 
The area of this triangle is {{{expr(1/2)*a*a=expr(1/2)a^2}}}

So the desired probability is  

{{{(AREA_OF_SMALLER_TRIANGLE)/(AREA_OF_LARGER_TRIANGLE)}}}

{{{ ((a^2)/8) /(a^2/2) }}}{{{""=""}}}{{{expr(a^2/8)}}}{{{""*""}}}{{{expr(2/a^2)}}}{{{""=""}}}{{{expr(cross(a^2)/8)}}}{{{""*""}}}{{{expr(2/cross(a^2))}}}{{{""=""}}}{{{2/8}}}{{{""=""}}}{{{1/4}}}

Edwin</pre>