$\displaystyle y = mx + b$
$\displaystyle A = \dfrac{xb}{2}$
$\displaystyle m = \dfrac{b-8}{-6}$
Good! Now, you need to find A' to get the minimum area. But for that, you'll need to get A in terms of either x alone, or b alone.
$\displaystyle y = \dfrac{8-b}{6} x + b$
At point (x, y) = (6, 8), find x in terms of b, or vice versa.
Then, substitute this into your expression for the area.
Differentiate the area, and solve for A' = 0.
You'll get a value of b or x depending on what you chose previously.
Substitute this value into the equation y = mx + b and use the point (6, 8) again to find the other value you need.