Question

Consider the family of lines $(x-y-6)+\lambda (2x+y+3)=0$ and $(x+2y-4)+\mu (3x-2y-4)=0.$ If the lines of these two families are at right angle to each other, then the locus of their point of intersection is

• A. $x^2+y^2+3x+4y-3=0$
• B. $x^2+y^2-3x+4y-3=0$
• C. $x^2+y^2=25$
• D. $x^2+y^2+6x+8y-3=0$

Answer 1

The correct option is B $x^2+y^2-3x+4y-3=0$


$L_1:(x-y-6)+\lambda (2x+y+3)=0$
Intersection point of these two lines
$x-y-6=0$ and $2x+y+3=0$ is $(1,-5)$
$L_2:(x+2y-4)+\mu (3x-2y-4)=0$
Intersection point of these two lines $x+2y-4=0$ and $3x-2y-4=0$ is $(2,1)$.
Since lines of these two families are perpendicular to each other,
$\therefore (1,-5)$ and $(2,1)$ will be the diametric end points of a circle.
Hence, $(x-1)(x-2)+(y+5)(y-1)=0$
$\Rightarrow x^2+y^2-3x+4y-3=0$ is the locus of their point of intersection.