A and B are the points of intersection of the line $4y = 2x + 1$ and the curve $3y - x = 4xy$.

1. Express $x$ in terms of $y$ using the linear equation.

From $4y = 2x + 1$, we have:
$2x = 4y - 1$
$x = \frac{4y - 1}{2}$

2. Substitute the expression for $x$ into the non-linear equation to form a quadratic equation in $y$.

Substitute $x = \frac{4y - 1}{2}$ into $3y - x = 4xy$:
$3y - \left(\frac{4y - 1}{2}\right) = 4\left(\frac{4y - 1}{2}\right)y$
Multiply the entire equation by 2 to clear the denominator:
$6y - (4y - 1) = 4y(4y - 1)$
$2y + 1 = 16y^2 - 4y$
$16y^2 - 6y - 1 = 0$

3. Solve the quadratic equation for $y$.

Factorising the quadratic equation:
$(8y + 1)(2y - 1) = 0$
$y = -\frac{1}{8}$ or $y = \frac{1}{2}$

4. Find the corresponding $x$ values and state the coordinates of A and B.

When $y = -\frac{1}{8}$:
$x = \frac{4(-\frac{1}{8}) - 1}{2} = \frac{-\frac{1}{2} - 1}{2} = -\frac{3}{4}$

When $y = \frac{1}{2}$:
$x = \frac{4(\frac{1}{2}) - 1}{2} = \frac{2 - 1}{2} = \frac{1}{2}$

The coordinates are $(-\frac{3}{4}, -\frac{1}{8})$ and $(\frac{1}{2}, \frac{1}{2})$.

1. Find the gradient of the line $AB$.

The equation of the line $AB$ is $4y = 2x + 1$.
Rearranging to the form $y = mx + c$:
$y = \frac{1}{2}x + \frac{1}{4}$
The gradient of $AB$, $m_1 = \frac{1}{2}$.

2. Find the gradient of Henry's line.

The equation of Henry's line is $2y - 4x = 5$.
Rearranging to the form $y = mx + c$:
$2y = 4x + 5$
$y = 2x + \frac{5}{2}$
The gradient of this line, $m_2 = 2$.

3. Check the condition for perpendicular lines and conclude.

For two lines to be perpendicular, the product of their gradients must be $-1$.
$m_1 \times m_2 = \frac{1}{2} \times 2 = 1$
Since $1 \neq -1$, the lines are not perpendicular. Therefore, Henry is incorrect.