If a variable line drawn through the intersection of the lines...

{L}_{1}: 4x + 3y - 12 = 0$<br /><br />${L}_{2}: 3x + 4y - 12 = 0$<br /><br />${L}_{1} + \lambda{L}_{2} = 0$<br /><br />$⇒\left(4x + 3y - 12\right) + \lambda\left(3x + 4y - 12\right) = 0$<br /><br />$⇒x\left(4 + 3\lambda\right) + y\left(3 + 4\lambda\right) - 12\left(1 + \lambda\right) = 0$<br /><br />Point$A\left(\dfrac{12\left(1 + \lambda\right)}{4 + 3\lambda}, 0\right)$<br /><br />Point$B\left(0, \dfrac{12\left(1 + \lambda\right)}{3 + 4\lambda}\right)$<br /><br />Mid point$⇒ h = \dfrac{6\left(1 + \lambda\right)}{4 + 3\lambda}$.......\left(i\right)$

$k = \dfrac{6\left(1 + \lambda\right)}{3 + 4\lambda}$ .......\left(ii\right)$<br /><br />Eliminate$\lambda$from equations \left(i\right)$ and \left(ii\right)$then<br /><br />$6\left(h + k\right) = 7hk$<br /><br />Hence, locus is<br /><br />$6\left(x + y\right) = 7xy