Mathematics - Point Question with Solution | TestHub

$L_1 :4x+3y-12=0$

$L_2:3x+4y-12=0$

$L_1+\lambda L_2=0$

$\Rightarrow$$\left(4x+3y-12\right)+\lambda \left(3x+4y-12\right)=0$

$\Rightarrow$$x\left(4+3\lambda \right)+y\left(3+4\lambda \right)-12\left(1+\lambda \right)=0$

Point$A\left(\frac{12\left(1+\lambda \right)}{4+3\lambda }, 0\right)$

Point$B\left(0,\frac{12\left(1+\lambda \right)}{3+4\lambda }\right)$

Mid point$\Rightarrow h=\frac{6\left(1+\lambda \right)}{4+3\lambda }$.......$\left(\mathrm{i}\right)$

$k=\frac{6\left(1+\lambda \right)}{3+4\lambda }$.......$\left(\mathrm{ii}\right)$

Eliminate$\lambda$from equations$\left(\mathrm{i}\right)$and$\left(\mathrm{ii}\right)$then

$6\left(h+k\right)= 7hk$

Hence, locus is

$6\left(x+y\right)= 7xy$