(12x2 $+$ $7xy-$ $12y^{2}\right)$ by $\left(3x+$ $+4y\right)$

$x^{2}+11xy-12y^{2}$

$2.$ Show that the four straight lines given by the equations, $12x^{2}+7xy-12y^{2}=0$ and $12x^{2}+7\times y$ $-12y^{2}-x+7y-1=0$ lie along sides of a square. ir point of