Let $A \equiv(0,0), B(3,0), C(0,-4)$ are vertices of $\triangle A B C$ then the co-ordinates of incentre of $\triangle A B C$ is

Given vertices $A(0,0)$, $B(3,0)$, and $C(0,-4)$, the side lengths are found using the distance formula.

Side $a$ opposite $A$ is $BC = \sqrt{(3-0)^2 + (0 - (-4))^2} = \sqrt{9 + 16} = 5$.

Side $b$ opposite $B$ is $AC = \sqrt{(0-0)^2 + (-4-0)^2} = 4$.

Side $c$ opposite $C$ is $AB = \sqrt{(3-0)^2 + (0-0)^2} = 3$.

The incenter coordinates are given by $I = \left( \frac{a x_1 + b x_2 + c x_3}{a + b + c}, \frac{a y_1 + b y_2 + c y_3}{a + b + c} \right)$.

$x = \frac{5 \cdot 0 + 4 \cdot 3 + 3 \cdot 0}{5 + 4 + 3} = \frac{12}{12} = 1$

$y = \frac{5 \cdot 0 + 4 \cdot 0 + 3 \cdot (-4)}{12} = \frac{-12}{12} = -1$

The incenter is $(1, -1)$, matching option B.