Family of Lines 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 which is farthest from $(1,-3)$.

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Any help with this problem would be really appreciated! 

We can write the equation as $\displaystyle (2\lambda+1)x+(1-\lambda)y+\lambda+1 =0$
Now Perpendicular Distance of Line from Point $\bf{P(1,-3)}$ is $ = PQ=\displaystyle \left|\frac{2\lambda+1+3\lambda-3+\lambda+1}{\sqrt{(2\lambda+1)^2+(1-\lambda)^2}}\right|$
Now $\displaystyle PQ = \left|\frac{6\lambda-1}{\sqrt{5\lambda^2+2\lambda+2}}\right|\;,$ Then we have to $\bf{Maximize}$ the value of $$\displaystyle \frac{6\lambda-1}{\sqrt{5\lambda^2+2\lambda+2}}$$
Now Let $$\displaystyle f(\lambda) = \frac{(6\lambda-1)^2}{5\lambda^2+2\lambda+2}$$.(Bcz If $f(\lambda)$ is $\bf{Max.}\;,$ Then $\left[f(\lambda)\right]^2$ is $\bf{Max.}$)
Now Let $$\displaystyle z = \frac{36\lambda^2+1-12\lambda}{5\lambda^2+2\lambda+2}\Rightarrow 5\lambda^2\cdot z+2\lambda \cdot z+2z = 36\lambda^2-12\lambda+1$$
So we get $$\displaystyle \left(5z-36\right)\lambda^2+2(z+6)\lambda+(2z-1) =0\;,$$ For real roots, Then $\bf{Discriminant\geq 0}$
So $$\displaystyle 4\left(z+6\right)^2-4\cdot (5z-36)\cdot (2z-1)\geq 0$$
So $(z+6)^2-(5z-36)\cdot (2z-1)\geq 0$
So $$\displaystyle z^2+36+12z-10z^2+77z-36\geq 0$$
So $$\displaystyle -9z^2+89z\geq 0\Rightarrow 9z^2-89z\leq 0$$
So we get $$\displaystyle z\leq \frac{89}{9}$$
So $$\displaystyle \bf{Max.(z) = \frac{89}{9}}$$
Now Put $\displaystyle z = \frac{89}{9}$ in $$\displaystyle z = \frac{(6\lambda-1)^2}{5\lambda^2+2\lambda+2}\;,$$ We Get $\displaystyle \lambda = -\frac{13}{11}$
