A triangle $ABC$ is right angled at $A$ has points A and $B$ as $(2, 3)$ and $(0,-1)$ respectively. If$BC= 5,$ then point $C$may be 

Slope of $AB=2\Rightarrow$slope of $AC=-\frac{1}{2}$

   $\Rightarrow \frac{y-3}{x-2}=-\frac{1}{2}\Rightarrow x+2y-8=0 ...\left(i\right)$

Also $x^2+(y+1{)}^2=25$

$\Rightarrow (8-2y{)}^2+(y+1{)}^2=25\left[\text{ from (i) }\right]$

$\Rightarrow y=2$ or $4$ and correspondingly $x = 4$ and $x=0.$ 

Hence, $C$ is $(0, 4) or (4, 2).$