Find the area of the triangle with the vertices (2,3), (4,1), (5,0)

The correct option is 0.

To find the area of a triangle with given vertices, we can use the coordinate geometry formula for the area of a triangle. Let the vertices of the triangle be:
$$(x_1,y_1)=(2,3)$$
$$(x_2,y_2)=(4,1)$$
$$(x_3,y_3)=(5,0)$$

The standard formula for the area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ is given by:
$$\text{Area}=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$$

Let us substitute the coordinate values of our vertices into this formula:
$$\text{Area}=\frac{1}{2}|2(1-0)+4(0-3)+5(3-1)|$$

Now, we simplify the terms inside the absolute value brackets step-by-step:
First term: $2(1-0)=2(1)=2$
Second term: $4(0-3)=4(-3)=-12$
Third term: $5(3-1)=5(2)=10$

Substitute these values back into the expression:
$$\text{Area}=\frac{1}{2}|2-12+10|$$

Add the terms inside the absolute value:
$$2-12+10=-10+10=0$$

Thus, the area becomes:
$$\text{Area}=\frac{1}{2}\left|0\right|=0$$

Since the area of the triangle is 0, it means that the three points (2,3), (4,1), and (5,0) are collinear (they lie on a single straight line) and do not form a proper triangle. Therefore, the area is 0 square units.