Question:

Step 1: Label the vertices. Let the vertices of the triangle be $A = (1, 3)$, $B = (0, 0)$, and $C = (2, 0)$. Step 2: Calculate the side lengths. The side lengths are calculated using the distance formula. Side $a$ (opposite to vertex $A$): $$a = |BC| = \sqrt{(2-0)^2 + (0-0)^2} = \sqrt{2^2 + 0^2} = \sqrt{4} = 2$$ Side $b$ (opposite to vertex $B$): $$b = |AC| = \sqrt{(2-1)^2 + (0-3)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$$ Side $c$ (opposite to vertex $C$): $$c = |AB| = \sqrt{(1-0)^2 + (3-0)^2} = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$$ Step 3: Apply the incentre formula. The coordinates of the incentre $I(x, y)$ are given by the formula: $$I = \left(\frac{a x_A + b x_B + c x_C}{a+b+c}, \frac{a y_A + b y_B + c y_C}{a+b+c}\right)$$ Substitute the calculated side lengths and vertex coordinates: $a=2$, $b=\sqrt{10}$, $c=\sqrt{10}$ $A=(1,3)$, $B=(0,0)$, $C=(2,0)$ Step 4: Calculate the x-coordinate. $$x = \frac{2(1) + \sqrt{10}(0) + \sqrt{10}(2)}{2 + \sqrt{10} + \sqrt{10}}$$ $$x = \frac{2 + 0 + 2\sqrt{10}}{2 + 2\sqrt{10}}$$ $$x = \frac{2 + 2\sqrt{10}}{2 + 2\sqrt{10}}$$ $$x = 1$$ Step 5: Calculate the y-coordinate. $$y = \frac{2(3) + \sqrt{10}(0) + \sqrt{10}(0)}{2 + \sqrt{10} + \sqrt{10}}$$ $$y = \frac{6 + 0 + 0}{2 + 2\sqrt{10}}$$ $$y = \frac{6}{2(1 + \sqrt{10})}$$ $$y = \frac{3}{1 + \sqrt{10}}$$ To rationalize the denominator, multiply the numerator and denominator by the conjugate $1 - \sqrt{10}$: $$y = \frac{3}{1 + \sqrt{10}} \cdot \frac{1 - \sqrt{10}}{1 - \sqrt{10}}$$ $$y = \frac{3(1 - \sqrt{10})}{1^2 - (\sqrt{10})^2}$$ $$y = \frac{3(1 - \sqrt{10})}{1 - 10}$$ $$y = \frac{3(1 - \sqrt{10})}{-9}$$ $$y = \frac{1 - \sqrt{10}}{-3}$$ $$y = \frac{\sqrt{10} - 1}{3}$$ Step 6: State the incentre coordinates. The incentre of the triangle with vertices $(1, 3)$, $(0, 0)$, and $(2, 0)$ is $\left(1, \frac{\sqrt{10} - 1}{3}\right)$.