Question

Let $L$ be a line passing through the point of intersection of the lines $x+2y+1=0$ and $2x+3y-1=0$. The locus of the circumcentre of the triangle formed by $L$ and coordinate axes is

• A. $6x-10y+4xy=0$
• B. $6x-10y-4xy=0$
• C. $6x+10y-4xy=0$
• D. $6x+10y+4xy=0$

Answer 1

The correct option is A $6x-10y+4xy=0$

Equation of line $L$ is
$(x+2y+1)+\lambda (2x+3y-1)=0\Rightarrow (1+2\lambda )x+(2+3\lambda )y+(1-\lambda )=0$
The triangle formed is rightangle so the cirmcumcentre is the midpoint of hypotenuse
Let the midpoint be $(h,k)$
Now, the intercepts of the line is
$x-\text{intercept is}\frac{\lambda -1}{1+2\lambda }y-\text{intercept is}\frac{\lambda -1}{2+3\lambda }$
Now, the midpoint is
$(h,k)=(\frac{\lambda -1}{2(1+2\lambda )},\frac{\lambda -1}{2(2+3\lambda )})\Rightarrow 2h=\frac{\lambda -1}{1+2\lambda }\Rightarrow \lambda =-\frac{2h+1}{4h-1}$
Similarly,
$\lambda =-\frac{4k+1}{6k-1}$
Now, the locus is
$\frac{2h+1}{4h-1}=\frac{4k+1}{6k-1}\Rightarrow 12hk-2h+6k-1=16hk+4h-4k-1\therefore 6x-10y+4xy=0$