Find the equation of the locus of the point P as it moves equidistant from the lines $x = 1$ and $y = 1$

Question

Find the equation of the locus of the point P as it moves equidistant from the lines $x = 1$ and $y = 1$

I cannot see where I am going wrong here.

I say,

$|x-1| = |y-1|$

$(x-1)^2 = (y-1)^2\implies x^2-y^2=2x-2y$

$\implies (x-y)(x+y)=2(x-y)\implies x+y-2=0$

but my books says the answer is $(x-y)(x+y-2) = 0$

Answer 1

Your book is right. From your penultimate step, we have \begin{align} &(x-y)(x+y) = 2(x-y)\\ \implies &(x-y)(x+y)-(x-y)(2)=0\\ \implies &(x-y)(x+y-2)=0 \end{align} I don't understand how you got $x+y+2=0$ in the last step.