Question 1200960
<pre>
The line {{{x/a + y/b = 1}}} is in intercept form with x-intercept (a,0) and 
y-intercept (0,b). So the midpoint of AB is {{{(matrix(1,3,(a+0)/2,",",(0+b)/2))}}} = {{{(matrix(1,3,a/2,",",b/2))}}}.

Since it lies on the line {{{2x + y = 10}}}, 

{{{2(a/2) + (b/2) = 10}}},

{{{a+b/2=10}}}
{{{2a+b=20}}}
{{{b=20-2a}}}

Since the distance AB = 10

{{{sqrt((a-0)^2+(0-b)^2)=10}}}
{{{sqrt(a^2+b^2)=10}}}
{{{a^2+b^2=100}}}

{{{a^2+(20-2a)^2=100}}}
{{{a^2+(400-40a+4a^2)=100}}}
{{{5a^2-40a=100}}}
{{{5a^2-40a-100=0}}}
{{{a^2-8a-20=0}}}
{{{(a+2)(a-20)=0}}}
a+2=0;  a-20=0
  a=-2;    a=20
{{{b=20-2a}}}
{{{b=20-2(-2)}}}, {{{b=20-2(20)}}}
{{{b=20+4}}},     {{{b=20-20}}}
{{{b=24}}},       {{{b=0}}}

So at first, there seem to be two answers. The first one

a=-2, b=24

{{{x/(-2) + y/24 = 1}}}
{{{-12x+y=24}}}
{{{y=12x+24}}}
with A=(a,0)=(-2,0), B=(0,24) 
and midpoint M of AB = {{{(matrix(1,3,a/2,",",b/2))}}} = {{{(matrix(1,3,(-2)/2,",",((24))/2))}}} = (-1,12)

Here is solution with a=-2, b=24. The green line is 2x + y = 10.
{{{drawing(320,560,-7,7,-5,30,

locate(-2.4,1.11,A), locate(0.33,24.6,B), locate(-1.6,12.4,M),
graph(320,560,-7,7,-5,30,24+12x,10-2x))}}} 

However what seemed like another solution

a=20, b=0

turned out to be extraneous because when we substitute a=20, b=0
in {{{x/a + y/b = 1}}}, we get

{{{x/20 + cross(y/0) = 1}}}

We have an undefined term in the equation, since division by 0 is
not permitted.

If the equation had first been multiplied through by ab so that we
would have had {{{bx+ay=ab}}} then we may have had a second solution, 
but since it was given as {{{x/a + y/b = 1}}}, the "second solution" 
was extraneous.  So there is just one solution a=-2, b=24.

Edwin</pre>