Question

The locus of point of intersection of the perpendicular lines one belonging to $(x+y-2)+\lambda (2x+3y-5)=0$ and other to $(2x+y-11)+\lambda (x+2y-13)=0$ is

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

Answer 1

Answer: (A)

$x^2+y^2-4x-6y+8=0$

Given family of lines are

$(x+y-2)+\lambda (2x+3y-5)=0$

Intersection point of $(x+y-2=0$ and $2x+3y-5=0$ is

$x=1,y=1$ i.e $(1,1)$

Also

$(2x+y-11)+\lambda (x+2y-13)=0$

Intersection of line $2x+y-11=0$ and $(x+2y-13)=0$ is

$x=3,y=5$ i.e.$(3,5)$

Let $(h,k)$ be the point of intersection

As the line are perpendicular

$\frac{k-1}{h-1}\times \frac{k-5}{h-3}=-1⟹k^2-6k+5=-(h^2-4h+3)⟹k^2+h^2-4h-6k+8=0$

Hence the locus of intersection is $x^2+y^2-4x-6y+8=0$.