Determine co-ordinates of point? 
How do you work out the co-ordinates of $P_{t+1}$ if you're given the co-ordinates of $P_0, P_1, P_2$ ? example (1,1);(2,5);(6,6)
The angle alpha remains constant between $(P_1, P_2)$ and $(P_2, P_{t+1})$

Slope of $P_0P_1 = \frac{1-5}{1-2} = 4$ 
Slope of $P_2P_1 = \frac{6-5}{6-2} = \frac{1}{4}$ 
Slope of $P_{t+1}P_2 = \frac{y-6}{x-6} = \tan \alpha$
$\alpha=\tan^{-1}\left(\frac{4-\frac{1}{4}}{1+4\cdot \frac{1}{4}}\right) = \tan^{-1}\left(\frac{\frac{15}{4}}{2}\right) = \tan^{-1}\left(\frac{15}{8}\right)$
By the problem, $\frac{15}{8}=\left(\frac{|\frac{y-6}{x-6}-\frac{1}{4}|}{1+\frac{y-6}{x-6}\frac{1}{4}}\right)=\left(\frac{|4y-24-x+12|}{4x-24+y-6}\right)=\left(\frac{|4y-x-12|}{4x+y-30}\right)$
or, $60x+15y-450 = |32y-8x-96|$
$$=>60x+15y-450 = 32y-8x-96$$
or, $$68x-17y-354=0$$
$$\mathbf{OR}$$
$$=>60x+15y-450 =-(32y-8x-96)$$
or,$$52x+47y=546=0$$
From both lines applying distance formula, you will get 4 points. You will reject 2 points on account of their lying on that part of the st. line where the lines make angle of $180-\alpha$. 
This pic is quite bad, still I was talking about 1,2,3 and 4. You have to reject 1 and 4. 

Hope you can do this much.
