<s>[INST] 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! 
 [/INST] Notice, the equation of straight lines $$x+y+\lambda(2x-y+1)=0$$  $$(2\lambda +1)x+(1-\lambda)y+\lambda=0$$  
Now, the distance say $D$ of the above line from the given point $(1, -3)$ is given as follows $$D=\left|\frac{(2\lambda +1)(1)+(1-\lambda)(-3)+\lambda}{\sqrt{(2\lambda +1)^2+(1-\lambda )^2}}\right|$$
$$=\left|\frac{(2\lambda +1)(1)+(1-\lambda)(-3)+\lambda}{\sqrt{(2\lambda +1)^2+(1-\lambda )^2}}\right|$$
$$=\left|\frac{6\lambda-2}{5\lambda^2+2\lambda+2}\right|$$
Now, considering the following function: $f(\lambda)=\frac{6\lambda-2}{5\lambda^2+2\lambda+2}$
$$\frac{df(\lambda)}{d\lambda}=\frac{d}{d\lambda}\left(\frac{6\lambda-2}{5\lambda^2+2\lambda+2}\right)$$ $$=\frac{(5\lambda^2+2\lambda+2)(6)-(6\lambda-2)(10\lambda+2)}{(5\lambda^2+2\lambda+2)^2}$$
$$=\frac{-30\lambda^2+20\lambda+16}{(5\lambda^2+2\lambda+2)^2}$$
Now, for maximum value of $D$ we have $$\frac{df(\lambda)}{d\lambda}=0$$
$$\frac{-30\lambda^2+20\lambda+16}{(5\lambda^2+2\lambda+2)^2}=0$$ 
$$15\lambda^2-10\lambda-8=0$$ $$\lambda=\frac{5\pm\sqrt{145}}{15}$$$$\iff \lambda =\frac{5+\sqrt{145}}{15}\ \vee\ \lambda=\frac{5-\sqrt{145}}{15}$$
Now, find $\frac{d^2f(\lambda)}{d\lambda^2}$ & substitute these values of $\lambda $ individually in $\frac{d^2f(\lambda)}{d\lambda^2}$ then you will find that for $\lambda=\frac{5+\sqrt{145}}{15}$,   $\frac{d^2f(\lambda)}{d\lambda^2}<0$
Thus, the function $f(\lambda)$ will be maximum & hence the distance $D$ of the given point from the line will be maximum at $\lambda=\frac{5+\sqrt{145}}{15}$
Now, substituting this value   of $\lambda $ in the given equation of family of straight lines we get the equation of the straight line farthest from the given point $(1, -3)$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x+y+\left(\frac{5+\sqrt{145}}{15}\right)(2x-y-1)=0}}$$
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