Calculate a point on the line at a specific distance . I have two points which make a line $l$ , lets say $(x_1,y_1) , (x_2,y_2)$ . I want a new point $(x_3,y_3)$ on the line $l$ at a distance $d$ from $(x_2,y_2)$ in the direction away from $(x_1,y_1)$ . How should i do this in one or two equation . 

Any point $p$ on your line can be written as 
$$
  p = (x_1, y_1) + \lambda \cdot (x_2 - x_1, y_2 - y_1) 
$$
for some $\lambda \in \mathbb R$, as these points form a line on which $(x_1, y_1)$ (corresponding to $\lambda = 0$) and $(x_2, y_2)$ (with $\lambda = 1$) lie. 
A point lies on the same side of $(x_1, y_1)$ as $(x_2, y_2)$ when $\lambda > 0$, and "behind" $(x_2, y_2)$ if $\lambda > 1$. The distance of a point on $l$ to $(x_2, y_2)$ is given by 
\begin{align*}
  d^2 &= \bigl((1-\lambda)x_1 + \lambda x_2 - x_2\bigr)^2 + \bigl((1-\lambda)y_1 + \lambda y_2 - y_2\bigr)^2\\
      &= (1-\lambda)^2(x_1 - x_2)^2 + (1-\lambda)^2(y_1 - y_2)^2\\
\iff d &= \left|1-\lambda\right| \cdot \bigl((x_1-x_2)^2 + (y_1-y_2)^2\bigr)^{1/2}
\end{align*}
So we want a $\lambda > 1$ (that is $|1-\lambda| = \lambda - 1$) at distance $d$, giving
$$
\lambda = 1 + \frac d{\bigl((x_1-x_2)^2 + (y_1-y_2)^2\bigr)^{1/2}}
$$
