
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Given a point $P(x_0, y_0)$ and a line $y = kx + b$, the distance $d$ from point $P$ to the line $y = kx + b$ can be calculated using the formula $d= \dfrac{|kx_0 - y_0 + b|}{\sqrt{1+k^2}}$.

For example: Find the distance from point $P(-2,1)$ to the line $y = x + 1$.

Solution: Since the line $y = x + 1$ can be transformed into $x - y + 1 = 0$, where $k = 1$, $b = 1$.

Therefore, the distance from point $P(-2,1)$ to the line $y = x + 1$ is $d= \dfrac{|1×(-2)-1+1|}{\sqrt{1+1^2}}= \dfrac{2}{\sqrt{2}}= \sqrt{2}$.

Based on the above materials, answer the following questions:

$(1)$ Find the distance from the origin to the line $y = x + 1$;

$(2)$ If the distance from point $P(1,1)$ to the line $y = x + b$ is $1$, find the value of $b$;

$(3)$ Given the distance $d$ between two parallel lines $y = -x + 1$ and $y = kx + b$ is $d$, and $\sqrt{2} \leqslant x \leqslant 2\sqrt{2}$, find the range of values for $b$.

Present the answer in LaTex format: \boxed{Your answer}
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