In the diagram A(-8;6); B; C; and D; are the vertices of a rhombus. The equation of BD is 3x−y=0. The diagonals of a rhombus intersect at K. 2.1 Calculate the perimeter of ABCD. Leave your answer in surd form 2.2 Determine the equation of diagonal AC in the form y=mx+c

$$AD = \sqrt{(3 - (-8))^2 + (9 - 6)^2}$$ $$AD = \sqrt{(3 + 8)^2 + (3)^2}$$ $$AD = \sqrt{(11)^2 + 3^2}$$ $$AD = \sqrt{121 + 9}$$ $$AD = \sqrt{130}$$

Perimeter $= 4 \times AD$

$$\text{Perimeter} = 4\sqrt{130}$$

The perimeter of ABCD is $\boxed{4\sqrt{130}}$.

Rearrange the equation $3x - y = 0$ to the form $y = mx + c$:

$$y = 3x$$

The gradient of BD, $m_{BD}$, is 3.

Since the diagonals are perpendicular, the product of their gradients is $-1$.

$$m_{AC} \times m_{BD} = -1$$ $$m_{AC} \times 3 = -1$$ $$m_{AC} = -\frac{1}{3}$$

Using the point-slope form $y - y_1 = m(x - x_1)$:

$$y - 6 = -\frac{1}{3}(x - (-8))$$ $$y - 6 = -\frac{1}{3}(x + 8)$$ $$y - 6 = -\frac{1}{3}x - \frac{8}{3}$$

Add 6 to both sides:

$$y = -\frac{1}{3}x - \frac{8}{3} + 6$$ $$y = -\frac{1}{3}x - \frac{8}{3} + \frac{18}{3}$$ $$y = -\frac{1}{3}x + \frac{10}{3}$$

The equation of diagonal AC is $\boxed{y = -\frac{1}{3}x + \frac{10}{3}}$.