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
Best Answer
First, note that all the lines that can be represented by your equation pass through the point (-2,0). Obviously it follows that the foot of the perpendicular to the line from (2,3) should be (-2,0)