Find the distance between A and B, AB,
[tex]AB=\sqrt{(3-3)^2+(-2-8)^2}=10[/tex]
Next, find the distance between B and C, BC,
[tex]BC=\sqrt{(t-3)^2+(1-2)^2}=\sqrt{(t-3)^2+1}[/tex]
The equation is,
[tex]10=2\sqrt{(t-3)^2+1}[/tex]
[tex]25=(t-3)^2+1[/tex]
[tex]t^2-3t+10=25[/tex]
[tex]t^2-3t-15=0[/tex]
Use quadratic formula to get possible values of t,
[tex]t_1=\frac{3+\sqrt{69}}{2}[/tex]
[tex]t_2=\frac{3-\sqrt{69}}{2}[/tex]
Hope this helps :)
