If $Q$ is the image of the point $P\left(1,1\right)$ with respect to the straight line $x+y+1=0$, then the length of the perpendicular drawn from $Q$ to the line $3x-4y+3=0$ is

Given,

$Q$ is the image of the point $P\left(1,1\right)$ with respect to the straight line $x+y+1=0$

So, point $Q$ is by

$\frac{x-1}{1}=\frac{y-1}{1}=-2\left(\frac{1+1+1}{1^2+1^2}\right)$

$\Rightarrow x=-2 \& y=-2$

So $Q\equiv \left(-2,-2\right)$

Now the length of the perpendicular drawn from $Q\left(-2,-2\right)$ to the line $3x-4y+3=0$ will be $\left|\frac{-2\times 3+4\times 2+3}{\sqrt{3^2+4^2}}\right|=\frac{5}{5}=1$