Find the equation of line that passes through (6,8)
and make with both axes a triangle with minimum area.
My try:
let the equation be y=mx+b
area = 1/2 xb
m=(b-8)/-6
then .....
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.
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.