Let's solve the problem step by step:

a) Calculate the gradient of line AB

The formula for the gradient $m$ between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ is: $m = \frac{y_2 - y_1}{x_2 - x_1}$

Substitute $A(0, 6)$ and $B(3, 0)$: $m = \frac{0 - 6}{3 - 0} = \frac{-6}{3} = -2$

Gradient of AB: $-2$

b) Find the gradient of a line perpendicular to AB

If two lines are perpendicular, the product of their gradients is $-1$. Let the gradient of the perpendicular line be $m_\perp$. Then: $m \cdot m_\perp = -1$ $-2 \cdot m_\perp = -1$ $m_\perp = \frac{1}{2}$

Gradient of the perpendicular line: $\frac{1}{2}$

c) Find the equation of the line passing through point A and perpendicular to AB

The equation of a line is given by: $y - y_1 = m(x - x_1)$ where $m$ is the gradient, and $(x_1, y_1)$ is a point on the line.

Here:

• Gradient $m_\perp = \frac{1}{2}$
• Point $A(0, 6)$

Substitute into the formula: $y - 6 = \frac{1}{2}(x - 0)$ $y - 6 = \frac{1}{2}x$ $y = \frac{1}{2}x + 6$

Equation of the perpendicular line: $y = \frac{1}{2}x + 6$

Final Answers:

a) Gradient of AB: $-2$
b) Gradient of a line perpendicular to AB: $\frac{1}{2}$
c) Equation of the perpendicular line: $y = \frac{1}{2}x + 6$

Would you like detailed steps for any part?
Here are 5 follow-up questions:

1. What does it mean geometrically for two lines to be perpendicular?
2. How can you verify the correctness of the equation $y = \frac{1}{2}x + 6$ on a graph?
3. What would the equation look like if the line passed through $B(3, 0)$ instead of $A(0, 6)$?
4. How would the gradient of AB change if point $B$ was at $(6, 0)$?
5. How can you find the intersection point of two lines algebraically?

Tip: When dealing with gradients, always double-check the points' coordinates to avoid sign errors.