0

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

Steblo
  • 1,153

1 Answers1

2

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.

Martund
  • 14,706
  • 2
  • 13
  • 30
  • Thank you. It was a mistake on my part. However, I am still confused why I was wrong to divide both sides of your first line by (x-y) to get x+y-2=0 – Steblo May 02 '21 at 09:37
  • @Steblo, because you can cancel a term from both sides if and only if that is non-zero. For example, if some number $x$ satisfies $4x=3x$, then cancelling $x$ both sides gives you $4=3$ which is absurd. The key lies in the fact that only $x=0$ satisfies this equation. – Martund May 02 '21 at 11:51
  • Btw, the example I've given above is the key to lots of proofs of $0=1$ you'll often find on social media :P – Martund May 02 '21 at 11:52
  • Thank you. All is clear now! – Steblo May 02 '21 at 12:42