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