Consider the family of lines $a(3x+4y+6)+b(x+y+2)=0$ Find the equation....... Question : 
Consider the family of lines $a(3x+4y+6)+b(x+y+2)=0$ Find the equation of the line of family situated at the greatest distance from the point P (2,3) 
Solution : 
The given equation can be written as $(3x+4y+6)+\lambda (x+y+2)=0$ 
$\Rightarrow x(3+\lambda)+y(4+\lambda)+6+2\lambda =0....(1)$
Distance of point P(2,3) from the above line (1) is given by 
D= $\frac{|2(3+\lambda)+3(4+\lambda)+6+2\lambda|}{\sqrt{(3+\lambda)^2+(4+\lambda)^2}}$
$\Rightarrow D = \frac{(24+7\lambda)^2}{(3+\lambda)^2+(4+\lambda)^2}$
Now how to maximize the aboved distance please suggest. Thanks

Find the common point of intersection of this family of lines. Here it is (-2,0). Now, given point is (2,3). Equation of line passing through both the points is : 3x + 6 =4y. You actually only need the slope of this line which is : 3/4 Line perpendicular to this line passing through the common point will be at the greatest distance. Slope of the required line: -4/3Equation of the line :-4/3{x-(-2)} = y-0After solving, equation is : 4x + 3y + 8 =0.
