Ytukyg

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 $(3x+4y+6)+lambda (x+y+2)=0$
$Rightarrow x(3+lambda)+y(4+lambda)+6+2lambda =0....(1)$

Distance of point P(2,3) from the above line (1) is given by

D= $frac{|2(3+lambda)+3(4+lambda)+6+2lambda|}{sqrt{(3+lambda)^2+(4+lambda)^2}}$

$Rightarrow D = frac{(24+7lambda)^2}{(3+lambda)^2+(4+lambda)^2}$

Now how to maximize the aboved distance please suggest. Thanks

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)

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.