Do babies recognize themselves in the mirror?

The easiest way to solve this problem is by finding the slope of the line that intersects both the points $(\lambda,\lambda-1), (\lambda^2+1,\lambda)$ and intersects the line $3x+y=6\lambda$ perpendicularly. Since the perpendicular line intersects both the points then its slope is:

$\frac{\lambda-(\lambda-1)}{\lambda^2+1-\lambda}$

The slope of the line mirror is -3 (since $y=6\lambda-3x$ ). Therefore the slope of the perpendicular line has to be 1/3 to be perpendicular with the mirror line (if the slope is $m$ then the slope at 90° wrt $m$ is $-1/m$ ).

$\frac{\lambda-(\lambda-1)}{\lambda^2+1-\lambda}=\frac{1}{3}$

$\lambda^2+1-\lambda=3 \rightarrow \lambda^2-\lambda-2=0$

The equation has two solutions $\lambda=-1,2$ .

If we replace $\lambda$ with $-1$ then we have the points $(-1,-2)$ and $(2,-1)$ which are both bellow the line $y=-3x-6$ . This cannot be a solution since the line has to be the mirror of the two points.

If we replace $\lambda$ with $2$ then the mirror line is $y=-3x+12$ . The new points are: $(2,1)$ which is over the mirror line and $(5,2)$ which is bellow the mirror line.

We already have a solution but we can double check that $\lambda$ is equal to 2 by verifying that the two distances of the points wrt the mirror line are equal.

The perpendicular line is $y=x/3+b$ . By replacing $x$ and $y$ with one of the two points we have that $b=1/3$ . The point of intersection of the mirror line and the perpendicular line can be calculated by the following system of equations: $y+3x=12\\ 3y-x=1$ The intersection point is $(7/2,3/2)$ . We now can check that the distance between the two points and the intersection point is equal:

$\sqrt{(7/2-2)^2+(3/2-1)^2}=\sqrt{(7/2-5)^2+(3/2-2)^2}$

$(3/2)^2+(1/2)^2=(-3/2)^2+(-1/2)^2$

$(3/2)^2+(1/2)^2=(3/2)^2+(1/2)^2$