Question

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

Answer 1

$(x-y-6)+\lambda (2x+y+3)=0$

For lines

$x-y-6=0$

$2x+y+3=0$ $x-y-6=0$

__________ $1-y-6=0$

$3x-3=0$ $y=-5$

$\therefore x=1$

$(x+2y-4)+4(3x-2y-y)=0$

For lines

$x+2y-4=0$ $x+2y-4=0$

$3x-2y-4=0$ $2+2y-4=0$

__________ $2y-2=0$

$4x-8=0$ $\therefore$ y = + 1

$\therefore x=2$

Let point of intersection be (h, k)

$\left(\frac{k+5}{n-1}\right)\left(\frac{k-1}{n-2}\right)=-1$

$\Rightarrow k^2+5k-k-5=-(h^2-3h+2)$

$\Rightarrow h^2+k^2-3h+4k-3=0$

Locus is

$x^2+y^2-3x+4y-3=0$