Mathematics - Straight Lines Question with Solution | TestHub

Let the lines be $y=m_{1}x,y=m_{2}x,y=m_{3}x$. Then $m$ satisfies $d{m}^3-3m^2+3m+3=0$, so $m_1{m}_2{m}_3=-\frac{3}{d}$. With $m_1{m}_2=-1,m_3=\frac{3}{d}$. Substituting $m=\frac{3}{d}:\frac{27}{d^2}-\frac{27}{d^2}+\frac{9}{d}+3=0$, so $d=-3$. The equation becomes $x^3+x^{2}y-x{y}^2-y^3=0$, i.e. $(x+y)\left(x^2-y^2\right)=0$. The slopes are $-1,1,-1$, so the third line has slope -1.