Suppose ∆ A B C is an isosceles triangle with ∠ C = 90 ° , A = 2 , 3 and B = 4 , 5 . Then…

Plotting the diagram of given value we have,

So, $AC=BC$ or $A{C}^2=B{C}^2$

$\Rightarrow {\left(x-2\right)}^2+{\left(y-3\right)}^2={\left(x-4\right)}^2+{\left(y-5\right)}^2$

$\Rightarrow 4-4x+9-6y=16-8x+25-10$

$\Rightarrow 4x+4y=28$

$\Rightarrow x+y=7 ...\left(1\right)$

Also $AC\perp BC$ we get

$\Rightarrow$ Slope of $AC\times$ slope $BC=-1$

$\frac{3-y}{2-x}\times \frac{5-y}{4-x}=-1 ...\left(2\right)$

$\Rightarrow \left(3-y\right)\left(5-y\right)=\left(x-4\right)\left(2-x\right)$

$\Rightarrow 15+y^2-8y=-x^2+6x-8$

$\Rightarrow x^2+y^2-8y-6x+23=0$

From equation (1) $y=7-x$ we get,

$\Rightarrow x^2+{\left(7-x\right)}^2-8\left(7-x\right)-6x+23=0$

$\Rightarrow x^2+x^2+49-14x-56+8x$$-6x+23=0$

$\Rightarrow 2x^2-12x+16=0$

$\Rightarrow x^2-6x+8=0$

$\Rightarrow \left(x-4\right)\left(x-2\right)=0$

$\Rightarrow x=2$ or $4$

So point $C\equiv \left(2,5\right)$ or $\left(4,3\right)$

Now centroid will be

$G=\left(\frac{2+2+4}{3},\frac{5+5+3}{3}\right)$$\equiv \left(\frac{8}{3},\frac{13}{3}\right)$

or $G=\left(\frac{4+2+4}{3},\frac{3+3+5}{3}\right)$$=\left(\frac{10}{3},\frac{11}{3}\right)$