In triangle $ABC$, the coordinates of two vertices are $A(-1 ; 1), \quad B(1 ; 2)$. The third vertex $C$ is on the $x$-axis. The centroid of the triangle is equidistant from the two coordinate axes.

What is the area of the triangle?
Since $C$ is on the $x$-axis, its coordinates are $C(c ; 0)$. According to the known relationship for the centroid coordinates,

$$
S_{x}=\frac{(-1+1+c)}{3}=\frac{c}{3}, \quad S_{y}=\frac{(1+2+0)}{3}=1
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_e096d7a29a08ffc9b106g-1.jpg?height=434&width=591&top_left_y=352&top_left_x=756)

Furthermore, we know that the centroid is equidistant from the $x$ and $y$ axes, so both of its coordinates are 1; therefore, the coordinates of $C$ are: $C(3 ; 0)$.

Let the projections of points $A$ and $B$ on the $x$-axis be $A'$ and $B'$, respectively. The area of the quadrilateral $A' A B C$ can be easily determined as the sum of the area of triangle $B B' C$ and the area of trapezoid $A' A B B'$. If we subtract the area of triangle $A' A C$ from this, we get the area of triangle $A B C$, which is

$$
\frac{2 \cdot 2}{2}+\frac{(2+1) \cdot 2}{2}-\frac{4 \cdot 1}{2}=3 \text { square units. }
$$

Remark. A point in the plane is equidistant from the two coordinate axes if it lies on the line $y=x$ or $y=-x$. We did not consider the latter case in our solution, because if $C$ had coordinates $(-3 ; 0)$, the points $A, B, C$ would be collinear, and thus would not form a triangle.