The distance from the origin to the image of $(1,1)$ with respect to the line $x+y+5=0$ is
The distance from the origin to the image of $(1,1)$ with respect to the line $x+y+5=0$ is
$7 \sqrt{2}$
$3 \sqrt{2}$
$6 \sqrt{2}$
$4 \sqrt{2}$
Solution
As we know that the image of $(1,1)$ with respect to line $x+y+5=0$ is
$\begin{aligned}
& \frac{x-1}{1}=\frac{y-1}{1}=\frac{2(1+1+5)}{1+1} \\
& \Rightarrow x-1=-7, y-1=-7 \\
& \Rightarrow x=-6, y=-6
\end{aligned}$
$\therefore \quad$ Image of point $(1,1)$ is $(-6,-6)$
Now, distance from origin
This is the required transformed equation.
$\begin{aligned}
& \mathrm{D}=\sqrt{(0+6)^2+(0+6)^2} \\
& \mathrm{D}=\sqrt{72}=6 \sqrt{2}
\end{aligned}$