Question

Is there a real value of $\lambda$ for which the image of the point $(\lambda ,\lambda -1)$ by the line mirror $3x+y=6\lambda$ is the point $({\lambda }^2+1,\lambda )$? If so find $\lambda$.

Answer 1

$3\left(\frac{{\lambda }^2+\lambda +1}{2}\right)+\left(\frac{2{\lambda }_1}{2}\right)=6\lambda$

$3{\lambda }^2+3\lambda +3+2\lambda -1=12\lambda$

$3{\lambda }^2-7\lambda +2=0$

$3\lambda -6\lambda -\lambda +2=0$

$3\lambda (\lambda -2)-1(\lambda -2)=0$

$(3\lambda -1)(\lambda -2)=0$

$\lambda =2$ $3\lambda -1=0$

$3\lambda =1$

$\lambda =\frac{1}{3}$

$\lambda =\frac{1}{3},2$