Answer:the triangle has (2,) and (0, 4) as its vertices for a maximum area
Step-by-step explanation:let P(x,0) be the x-intercept and Q(0,y) be the y-intercept
label (1,2) as A(1,2)
the slope AP = slope AQ
-2/(x-1) = (2-y)/1
2x - xy - 2 + y = -2
y(1-x) = -2x
y = -2x/(1-x) or 2x/(x-1)
H^2 = x^2 + y^2
= x^2 + ( (2x/(x-1) )^2 --- simplify if needed
(don't know why we are finding the hypotenuse ? )
Area of triangle OPQ
= (1/2)base x height
= (1/2)xy
= (1/2)x(2x/(x-1))
= (1/2) (2x^2/(x-1) )
dA/dx = (1/2) [ (x-1)(4x) - 2x^2(1) ]/(x-1)^2
= (1/2) [ 4x^2 - 4x - 2x^2 ]/(x-1)^2
= 0 for a max of A
2x^2 - 4x = 0
x(2x-4) = 0
x=0 ---- > makes no sense, look at diagram
or
x = 2
then y = 2(2)/(2-1) = 4
the triangle has (2,) and (0, 4) as its vertices for a maximum area
