Two perpendicular lines

Let the slope of the lines be m & -1/m. then the equation of lines would be y = mx +3 & my +x = 3m respectively. And their point of intersection would be -3/m and 3m respectively. So area of triangle (Let it be A) = b h/2 = [3m +3/m] 3/2. or 2Am = 9m^{2} +9 or 9m^{2} - 2Am +9=0 Since this quad. equation should have 2 solutions. So its D > 0. Applying this we would get A > 9. So A = 8 or 6 is not possible.

Let x be the point on the the right of the y-axis where one of the two perpendicular lines cuts the x-axis. Then some thought involving similar triangles shows that the area under the triangle described in the problem will be : $\frac { 3 }{ 2 } \left( x+\frac { 9 }{ x } \right)$ I took derivatives and solved symbolically using the Python sympy library :

>>> from sympy import *

>>> x=symbols('x')

>>> diff(1.5*(x+9.0/x),x)

1.5 - 13.5/x**2

>>> solve(Eq(_),x)

[-3.00000000000000, 3.00000000000000]

>>> (1.5*(x+9.0/x)).subs(x,3)

9.00000000000000

The minimum area is 9.

Area of triangle =1/2 h b h=3 b=x2+x2

Where x1 and X2 are x coordinates where two perpendicular lines cut (3/x1) (3/x2)=1 [two perpendicular lines have slop m1 m2=-1] 9/x1 x2=1 We have area 3/2 (x1+x2)=a From above we have x2=9/x1 So eq is 3/2 (x1+9/x1)=a It is quadratic eq 3x²-2 a x+27=0 To be x real b²-4 a c>=0 a² 4-4 3 27>=0 a²>=81 a>9 So area must be greater than or equal to 9