The locus of the point which is equidistant from the point $(1,1)$ and the line $\mathrm{x}+\mathrm{y}+1=0$ is
The locus of the point which is equidistant from the point $(1,1)$ and the line $\mathrm{x}+\mathrm{y}+1=0$ is
$x^2-y^2+6 x+4 y-3=0$
$(x-y)^2-6(x+y)+3=0$
$(x+y)^2+6(x-y)+3=0$
$x^2+y^2-2 x-2 y+4=0$
Solution
Let $\left(x_1, y_1\right)$ be the point which is at a distance $d$ from the $(1,1)$ then we have:
$\sqrt{\left(x_1-1\right)^2+\left(y_1-1\right)^2}=d \Rightarrow\left(x_1-1\right)^2+\left(y_1-1\right)^2=d^2...(i)$
$\because$ Point $\left(x_1, y_1\right)$ is at the same distance from the line
$\Rightarrow d=\frac{x_1+y_1+1}{\sqrt{1+1}}=\frac{x_1+y_1+1}{\sqrt{2}}...(ii)$
From equations (i) \& (ii)
$\begin{aligned}
& \left(x_1-1\right)^2+\left(y_1-1\right)^2=\frac{1}{2}\left(x_1+y_1+1\right)^2 \\
& \Rightarrow 2\left[x_1^2+1-2 x_1+y_1^2+1-2 y_1\right] \\
& =x_1^2+y_1^2+1+2 x_1 y_1+2 x_1+2 y_1 \\
& \Rightarrow x_1^2+y_1^2-2 x_1 y_1-6 x_1-6 y_1+3=0 \\
& \Rightarrow\left(x_1-y_1\right)^2-6\left(x_1+y_1\right)+3=0
\end{aligned}$
Taking locus of point $\left(x_1, y_1\right)$, we get:
$(x-y)^2-6(x-y)+3=0$