Given the line L with equation y = 2x + 3, and points B and C on L such that A is the midpoint of BC. If D = (1, 0) and BD is perpendicular to L, find the equation of line BD.

Point A is the y-intercept of the line $y = 2x + 3$. To find the y-intercept, we set $x = 0$.

The coordinates of point A are $(0, 3)$.

Points B and C are on the line $y = 2x + 3$. The condition $AB = AC$ means that point A is the midpoint of the line segment BC, because A, B, and C are collinear.

The line AC is the line $y = 2x + 3$. The slope of this line, $m_{AC}$, is the coefficient of $x$.

Line BD is perpendicular to line AC. The product of the slopes of two perpendicular lines is $-1$.
Let $m_{BD}$ be the slope of line BD.

Line BD passes through point D$(1, 0)$ and has a slope $m_{BD} = -\frac{1}{2}$. Using the point-slope form $y - y_1 = m(x - x_1)$:

The equation of BD is $\boxed{y = -\frac{1}{2}x + \frac{1}{2}}$.

Point B is the intersection of the line $y = 2x + 3$ and the line $y = -\frac{1}{2}x + \frac{1}{2}$. We set the y-values equal to find the x-coordinate:

Multiply the entire equation by 2 to clear fractions:

Substitute $x = -1$ into the equation $y = 2x + 3$ to find the y-coordinate:

The coordinates of point B are $\boxed{(-1, 1)}$.

Since A$(0, 3)$ is the midpoint of BC, and B is $(-1, 1)$, we use the midpoint formula:

For the x-coordinate:

For the y-coordinate:

The coordinates of point C are $\boxed{(1, 5)}$.