A family of lines is given by $$\left( {1 + 2\lambda } \right)x + \left( {1 - \lambda } \right)y + \lambda = 0,\,\lambda $$       being the parameter. The line belonging to this family at the maximum distance from the point $$\left( {1,\,4} \right)$$  is :

Solution :

Here $$x + y + \lambda \left( {2x - y + 1} \right) = 0$$
$$ \Rightarrow $$ each line passes through the point of intersection of the lines $$x+y=0$$   and $$2x-y+1=0,$$    which is $$\left( { - \frac{1}{3},\,\frac{1}{3}} \right)$$
The required line passes through $$\left( { - \frac{1}{3},\,\frac{1}{3}} \right)$$   and is perpendicular to the line joining $$\left( {1,\,4} \right)$$  and $$\left( { - \frac{1}{3},\,\frac{1}{3}} \right).$$
$$\therefore \,'m'$$  of the required line $$ = \frac{{ - 1}}{{\frac{{4 - \frac{1}{3}}}{{1 + \frac{1}{3}}}}} = - \frac{4}{{11}}$$
$$\therefore $$ the required line is $$y - \frac{1}{3} = - \frac{4}{{11}}\left( {x + \frac{1}{3}} \right){\text{ or }}12x + 33y = 7$$