1. ## Cartesian Plane

Find an equation of the set of points in the Cartesian plane that are equidistant from the point (-1,1) and the line y=x.

Find an equation of the set of points in the Cartesian plane
that are equidistant from the point $(-1,1)$ and the line $y=x.$
Code:
              |
| P
| *(x,y)      *
/  \        *
A  / |   \     *
*   |    \  *
(-1,1) |     *
|   *  B
| *
- - - - * - - - - - - - -
* |
*   |

The distance from point $P(x,y)$ to point $A(-1,1)$

. . is given by: . $PA \;=\;\sqrt{(x+1)^2 + (y-1)^2}$

The distance from a point $P(x_1,y_1)$ to a line $ax + by + c\:=\:0$

. . is given by: . $d \;=\;\frac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}$

Hence, the distance from $P(x,y)$ to the line $x - y \:=\:0$

. . is given by: . $PA \;=\;\frac{|x-y|}{\sqrt{1^2 + (-1)^2}} \;=-\;\frac{|x-y|}{\sqrt{2}}$

Since $PA = PB$, we have: . $\sqrt{(x+1)^2 + (y-1)^2} \;=\;\frac{|x-y|}{\sqrt{2}}$

Now simplify . . .