Answer
$(\dfrac{2a}{3}, \dfrac{b}{3})$
Work Step by Step
The area of a triangular region can be computed as: $A=\dfrac{1}{2} \times ab$
The $x$ -coordinate of the center of mass can be computed as: $$\overline{x}=\dfrac{1}{2A} \oint_{C} x^2 dy \\= \dfrac{1}{ab} \int_0^a \int_0^{bx/a} 2x dy dx\\=\dfrac{1}{ab} \times \int_0^a [2xy]_0^{bx/a} \\ =\dfrac{2a}{3}$$
The $y$ -coordinate of the center of mass can be computed as $$\overline{y}=-\dfrac{1}{2A} \oint_{C} x^2 dy \\= -\dfrac{1}{ab} \int_0^a \int_0^{bx/a} 2y dy dx\\=\dfrac{b}{a^3} \times \int_0^a x^2 dx \\ =\dfrac{b}{3}$$
So, the center of mass is: $(\dfrac{2a}{3}, \dfrac{b}{3})$
