Family of Straight line : Consider a family of straight lines $(x+y) +\lambda (2x-y +1) =0$. Find the equation of the straight .... Problem : Consider a family of straight lines $(x+y) +\lambda ( 2x-y +1) =0$.
Find the equation of the straight line belonging to this family that is farthest from $(1,-3)$. 
Solution:
Let the point of intersection of the family of lines be P. If solve :
$$\left\{\begin{matrix}
x+y=0 & \\ 
2x-y+1=0 & 
\end{matrix}\right.$$
We get the point of intersection which is $P \left(-\dfrac{1}{3}, \dfrac{1}{3} \right)$ 
Now let us denote the point $(1,-3)$ as $Q$. So, now how to find $\lambda$ so that this will be fartheset from $Q$. 
If we see the slope of $PQ = m_{PQ} = -\dfrac{5}{2}$ 
Any line perpendicular to $PQ$ will have slope $\dfrac{2}{5}$ Please suggest further.. thanks.

HINT:
We can rewrite the equation as $$x(1+2\lambda)+y(1-\lambda)+\lambda=0$$
If $d$ is the perpendicular distance from $(1,-3)$  
$$d^2=\frac{\{1(1+2\lambda)+(-3)(1-\lambda)+\lambda\}^2}{(1+2\lambda)^2+(1-\lambda)^2}$$
We need to maximize this which can be done using the pattern described here or here
