Let's solve the problem step by step.

Step 1: Find the coordinates of points A and B

The equation of the line is given by: $3x + 4y - p = 0$ To find the x-intercept (point $A$), set $y = 0$: $3x - p = 0 \implies x = \frac{p}{3}$ So, the coordinates of $A$ are $\left(\frac{p}{3}, 0\right)$.

To find the y-intercept (point $B$), set $x = 0$: $4y - p = 0 \implies y = \frac{p}{4}$ So, the coordinates of $B$ are $\left(0, \frac{p}{4}\right)$.

Step 2: Write the coordinates of $O$ and $C$

• $O$ is the origin, so $O = (0, 0)$.
• $C = (0, 8)$.

Step 3: Use the condition $AH = OH$

The orthocentre $H$ is the point where the altitudes of the triangle $ABC$ intersect. Since $AH = OH$, this implies that $H$ lies on the perpendicular bisector of $OA$.

Step 4: Equation of the perpendicular bisector of $OA$

The coordinates of $A$ are $\left(\frac{p}{3}, 0\right)$ and the coordinates of $O$ are $(0, 0)$. The midpoint $M$ of $OA$ is: $M = \left(\frac{\frac{p}{3} + 0}{2}, \frac{0 + 0}{2}\right) = \left(\frac{p}{6}, 0\right)$ The slope of $OA$ is 0, so the slope of the perpendicular bisector of $OA$ is undefined (a vertical line).

Thus, the equation of the perpendicular bisector of $OA$ is: $x = \frac{p}{6}$

Step 5: Determine the coordinates of $H$

Since $H$ lies on the perpendicular bisector $x = \frac{p}{6}$, $H$ has coordinates $\left(\frac{p}{6}, y_H\right)$.

Step 6: Use the condition $AH = OH$

For $AH = OH$, the distance from $A$ to $H$ must equal the distance from $O$ to $H$.

$AH = \sqrt{\left(\frac{p}{6} - \frac{p}{3}\right)^2 + (y_H - 0)^2}$ $OH = \sqrt{\left(\frac{p}{6} - 0\right)^2 + (y_H - 0)^2}$

Since $AH = OH$, the distance equation becomes: $\sqrt{\left(-\frac{p}{6}\right)^2 + y_H^2} = \sqrt{\left(\frac{p}{6}\right)^2 + y_H^2}$

Squaring both sides: $\frac{p^2}{36} + y_H^2 = \frac{p^2}{36} + y_H^2$ This simplifies to: $\text{This equation is always true, hence no new information is provided by it.}$

Given that this didn't simplify further, we re-check our approach:

The relationship $AH = OH$ suggests symmetry, meaning that $H$ lies on the y-axis due to the equal distance requirement. So, $H$ is at the point where the perpendicular from $C$ intersects the y-axis.

Step 7: Equation of $AC$

The slope of $AC$ is: $\text{slope of } AC = \frac{8 - 0}{0 - \frac{p}{3}} = -\frac{24}{p}$

The equation of the line $AC$ is then: $y - 8 = -\frac{24}{p}(x - 0)$

Simplifying, we get: $y = -\frac{24}{p}x + 8$

Thus, the equation of line $AC$ is: $\boxed{24x + py - 8p = 0}$

This is the required equation of line $AC$.

Do you want any further details or have any questions? Here are some related questions you might be interested in:

1. How do you find the orthocenter of a triangle with given coordinates?
2. What is the significance of the perpendicular bisector in a triangle?
3. How do you find the equation of a line given a point and a slope?
4. What are some properties of triangles with equal altitudes from a point?
5. How can we use coordinate geometry to solve triangle problems?
6. What is the role of symmetry in determining geometric properties?
7. How can you derive the equation of a line in standard form from point-slope form?
8. How do altitudes relate to the sides of a triangle in orthocentric systems?

Tip: When solving geometry problems, always sketch the scenario to visualize relationships and simplify problem-solving.