Question 389312
The slope from the point B(-1,-1) to C (-2,2) is (2--1)/(-2--1) = 3/(-1) = -3.  The equation of the line is y--1 = -3(x--1) ----->y+1 = -3(x+1) ----->y = -3x-3 - 1 -----> y = -3x-4.
Now the find the equation of the line perpendicular to the line y = -3x-4 and passing through A(9,9).  The slope of that line must be 1/3 (the negative reciprocal of -3).  Then the equation of the line is  y - 9 = (x-9)/3.
----> 3y -  27 = x - 9 ---->  3y = x + 18.
We now have to solve the system
 y = -3x-4
 3y = x + 18
Substitute the top equation into the bottom equation:
3(-3x-4) = x + 18;
-9x-12 = x + 18;
-30 = 10x;
x = -3.------>y = -3*-3 - 4 = 9-4 = 5.
Therefore segment AD is minimum for the point D(-3,5).