Minimising the area of triangle.

Question

Triangle $ABC$ has coordinate $A(2,1)$.

$B$ passes through the line $x=y$.

$C$ passes through the line $y=0$.

What is the minimum possible area of the triangle and what are its coordinates?

My attempt :

Let the coordinates be $A(2,1)$, $B(y, y)$, $C(x, 0)$.

Calculating the area it turns out to be

$\frac{1}{2}|{x+y-xy}|$.

Now I know that if we take the derivative and equate it to 0 then we can proceed further, but I can't find a way to relate x and y. I used that the sum of two sides is always greater than the third side but it was of no use.

Can anyone tell me how to proceed further. Thanks in advance!

Edit : The points are only integral

They are not on the same line.

Sorry, to those who have answered the question already, for missing out this information.

Answer 1

Note the following :

For the choice $x=0$ and $y=\epsilon$ , an arbitrarily small positive real number, the area can be made arbitrarily small.

Hence the minimal possible area of the triangle doesn't exist.

Suggestion : In case you consider only integral points, try computing the area considering few nearby points of $(2,1)$.

It won't be a difficult exercise.

Answer 2

Take $B(3,3)$ and $C(0,-3)$.

We see that $A,$ $B$ and $C$ are placed on the same line.

Id est, the minimum does not exist and the infimum is equal to $0$.

Answer 3

Continuing your work Area$= \dfrac12 |x+y-xy|$

$= \dfrac12 |1-(x-1)(y-1)|$

Since $x,$$y$ are integers $(x-1)(y-1)$ $= K $ Where $K$ is a integer, Therefore Minumum of $|1-K|$ will be $1$. Thus minimum Area $= 0.5$

Note that infinite triangles following the given conditions are possible with area $0.5$.