Analytical geometry

I don't believe this is  possible, geoNewbie

The  hypotenuse  will  be  the  fixed distance   from O to  A =   sqrt (10^2 + 10^2)  = sqrt (200)  =OA

Call the point  that we seek  ( x, x+ 6)

The distance  from    O to Q =    sqrt  [ ( x^2  + ( x + 6)^2  ]  =  OQ

And the distance  from   A to Q  = sqrt  [ ( x - 10) +  ( x + 6  - - 10)^2 ]  =

sqrt  [ (x - 10)^2  + (x + 16)^2 ]  =   AQ

Using the Law of Cosines   we  have that

OA^2   =  OQ^2  + AQ^2  - 2( OQ  * AQ)  cos (90)

The  cos 90   =  0     so  this boils down to

OA^2 =  OQ^2  + AQ^2

200   = ( x^2  + (x + 6)^2  +  ( x - 10)^2  + ( x + 16)^2

Using WolframAlpha to solve this, we get  no real solutions  for  x

Also.....playing around with this triangle in Geogebra, I  couldn't find any  angles   of OQA even approaching 90°

In fact  about   the  largest angle I could find is  ≈  38.xxx °

I wonder if  you might not mean  angle  QOA   is the right angle   ????

I believe you mean y = -x + 6, since this is worded exactly like an alcumus problem I saw a few days ago.

Anyway, the solution I did was:

The slope of the line from the origin to the point Q must be perpendicular to the slope of the line from (10, -10) to the point Q.

The equation for that is: $\frac{x}{y} = - \frac{-10-y}{10-x}$

Replace y with -x + 6 to solve for x:

$\frac{x}{-x+6} = -\frac{-10-(-x+6)}{10-x}\\\frac{x}{-x+6} = \frac{16-x}{10-x} \\ 10x-x^2=x^2-22x+96\\ 2x^2-32x+96=0\\x^2-16x+48=0\\(x-4)(x-12)=0\\x=4, x=12$

The y values corresponding to these x values are $y = 2, y = -6$, respectively.

Therefore, the answer to this question is $4+12+2-6=\boxed{12}$

https://www.desmos.com/calculator/efz9crqvir (this will help you visualize it)

Yes, I am sorry I had it wrong and I think I wasted everyone's time. I was struggling with it and thought I was missing something, and now I know what it was the "-" sign.