The problem asks to find the distance between the point $(-2, 3)$ and the line $y = 2x + 1$.

1. Problem Description

The problem asks to find the distance between the point $(-2, 3)$ and the line $y = 2x + 1$.

2. Solution Steps

First, we rewrite the equation of the line in the general form $Ax + By + C = 0$.

Given $y = 2x + 1$, we subtract $y$ from both sides to get $0 = 2x - y + 1$.

Thus, $A = 2$, $B = -1$, and $C = 1$.

The point is $(x_0, y_0) = (-2, 3)$.

The formula for the distance $d$ between a point $(x_0, y_0)$ and a line $Ax + By + C = 0$ is given by:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

Now, we plug in the values $A = 2$, $B = -1$, $C = 1$, $x_0 = -2$, and $y_0 = 3$ into the formula:

$d = \frac{|2(-2) + (-1)(3) + 1|}{\sqrt{2^2 + (-1)^2}}$

$d = \frac{|-4 - 3 + 1|}{\sqrt{4 + 1}}$

$d = \frac{|-6|}{\sqrt{5}}$

$d = \frac{6}{\sqrt{5}}$

To rationalize the denominator, we multiply the numerator and denominator by $\sqrt{5}$:

$d = \frac{6\sqrt{5}}{5}$

3. Final Answer

The distance between the point $(-2, 3)$ and the line $y = 2x + 1$ is $\frac{6\sqrt{5}}{5}$.