How to get Point between two points at any specific distance? I have two points, 
approximately we take values for that:
Point $A = (50, 150)$;
Point $B = (150, 50)$;
So the distance should be calculated here,
$\text{distance} = \sqrt{(B_x - A_x)(B_x - A_x) + (B_y - A_y)(B_y - A_y)}$;
Now I want any one poins which is far from Second point B at specific distance (Example, 10).
            B(x,y)
           /
          /
         C(x,y)
        /
       /
      /
     /
    /  
   /
  A(x,y)

Point c on Line Segment and its specific distance from point B(Ex, 10).. 
Which formula would be better to calculate C point here ?
Please help me about that.
 A: This what I come up with:
Find $(x_0,y_0)$ so that $10 = \sqrt{(50 - y_0)^2 + (150 - x_0)^2}$ and $(x_0,y_0)$ also lies on the line $y = 200 - x$.
Since $(x_0,y_0)$ lies on that line, we can write $y_0 = 200 - x_0$, so the distance formula becomes:
$10 = \sqrt{(-150 + x_0)^2 + (150 - x_0)^2} = \pm\sqrt{2}(x_0 - 150)$
Thus $x_0 = 150 \pm \frac{10}{\sqrt{2}}$, leading to:
$y_0 = 50 \mp \frac{10}{\sqrt{2}}$
A: I hope I have understood your question.
The general form of the line is $\lambda A + (1-\lambda) B$. You wish to find $\lambda$ so that the $y$-component is $10$.
Expanding gives: $\lambda A + (1-\lambda) B = (150-100 \lambda, 50+100 \lambda)$. Equating the $2$nd component to $10$ and solving for $\lambda$ gives $\lambda = -0.4$, from which we get the point $(190,10)$.
I think I misunderstood your question. If you wish to find points on the line at a specific distance $\delta$ from $B$, then you need to find the $\lambda$ that satisfies $||\lambda A + (1-\lambda) B -B || = |\lambda|\,||A-B|| = \delta$. Specifically, this gives $\lambda = \pm \frac{\delta}{||A-B||}$.
In this case, you have $\delta = 10$, and $||A-B|| = 100 \sqrt{2}$, so $\lambda = \pm \frac{1}{10\sqrt{2}}$. Substituting the positive value (which corresponds to the point between $A$ and $B$) in gives:
$$\lambda A + (1-\lambda) B = (150-\frac{10}{\sqrt{2}}, 50+\frac{10}{\sqrt{2}}).$$
The general formula for a point $\delta$ away from $B$ will be, of course:
$$(x,y) = (150\pm\frac{\delta}{\sqrt{2}}, 50 \mp\frac{\delta}{\sqrt{2}}).$$
A: Another easy to understand solution using vector arithmetic:
$$
\vec{a} = \begin{pmatrix}50\\150\end{pmatrix}, \vec{b}= \begin{pmatrix}150\\50\end{pmatrix} 
$$
Calculate direction vector $\vec{d}$ (the normalized distance vector between a and b):
$$
\vec{d} = \frac{\vec{a} - \vec{b}}{|\vec{a} - \vec{b}|}
$$
$\vec{d}$ has length 1 now. So to place a point $\vec{c}$ between $\vec{a}$ and $\vec{b}$ at a distance $x$ from $\vec{b}$, simply do:
$$
\vec{c} = \vec{b} + x \vec{d}
$$
