Question

Let P = (a,b) lie in the first quadrant. Find the slope of the line through P such that the triangle bounded by this line and the axes in the first quadrant has minimal area. Then show that P is the midpoint of the hypotenuse of this triangle.

Answer 1

use: y = m(x-a) + b

area = xy/2
= x(m(x-a) + b)/2

area' = [ m(x-a) + b + mx ]/2 = 0

m(x-a) + b + mx = 0 and solve

Answer 2

m(2x - a) + b = 0

m = -b/(2x-a)

Answer 3

y = -b(x-a)/(2x-a) + b

y = b [ 1 - (x-a)/(2x-a) ]

y = b (2x - a - x +a)/(2x-a)

y = bx/(2x-a)

midpoint eh ....

Answer 4

intercepts i spose would be needed