What does the equation of the locus of the points of the plane $P(x,y)$ whose distance from point $A(1,1)$ is equal to its distance from point $B(2,-2)$ correspond to?
The answer is $2x-6y-6=0$, but I can't figure out how to get there. As shown in the image, I tried to use the circumference equation, but it seems that it's much simpler. I also can't figure out how this is represented in the plane $P(x,y)$.
Attached an image; it's in spanish.
On
HINT
Consider the direction $v = B - A = (2,-2) - (1,1) = (1,-3)$ and the midpoint $M = (3/2,-1/2)$.
Hence the perpendicular direction to $v$ is parallel to $w = (-3,1)$, for example.
Hence the locus is given by the following line (why?): \begin{align*} (x,y) & = M + tw\\\\ & = (3/2,-1/2) + t(-3,1)\\\\ & = (3/2 - 3t,-1/2 + t) \end{align*}
then eliminate the variable $t$.
Can you take it from here?
Let $p = $ distance from $(x, y)$ to $(1, 1)$ and $q =$ distance from $(x, y)$ to $(2, -2)$.
It follows from the Pythagorean theorem that $p^2 = (x-1)^2 + (y-1)^2$ and $q^2 = (x-2)^2 + (y+2)^2$.
Hence, we have
$$p = q$$ $$\implies p^2 = q^2$$ $$\implies (x-1)^2 + (y-1)^2 = (x-2)^2 + (y+2)^2$$ $$\implies x^2 - 2x + 1 - x^2 + 4x - 4 = y^2 + 4y + 4 - y^2 + 2y - 1$$ $$\implies 2x - 3 = 6y + 3$$ $$\implies \fbox{$2x - 6y - 6 = 0$}$$