Question:

Step 1: Find the intersection point of the diagonals. The diagonals of a parallelogram bisect each other at a common point. We need to find where $AC$ and $BD$ intersect by solving the system: $$x + 2y = 3$$ $$2x + y = 3$$ Subtracting the first equation from the second: $$x - y = 0 \implies x = y$$ Substituting $x = y$ into the first equation: $$x + 2x = 3 \implies 3x = 3 \implies x = 1$$ Therefore, the center of the parallelogram is $O = (1, 1)$. Step 2: Find the angle between the diagonals. For line $AC: x + 2y - 3 = 0$, the normal vector is $(1, 2)$, so a direction vector is $(2, -1)$. For line $BD: 2x + y - 3 = 0$, the normal vector is $(2, 1)$, so a direction vector is $(1, -2)$. The cosine of the angle $\theta$ between the diagonals is: $$\cos\theta = \frac{|(2)(1) + (-1)(-2)|}{\sqrt{2^2 + (-1)^2} \cdot \sqrt{1^2 + (-2)^2}} = \frac{|2 + 2|}{\sqrt{5} \cdot \sqrt{5}} = \frac{4}{5}$$ Using the Pythagorean identity: $$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$$ Step 3: Apply the area formula for a parallelogram in terms of its diagonals. The area of a parallelogram can be expressed as: $$\text{Area} = \frac{1}{2}d_1 d_2 \sin\theta$$ where $d_1$ and $d_2$ are the lengths of the diagonals and $\theta$ is the angle between them. Given that $d_1 = AC = 4$ units, Area $= 8$ square units, and $\sin\theta = \frac{3}{5}$: $$8 = \frac{1}{2} \cdot 4 \cdot d_2 \cdot \frac{3}{5}$$ Step 4: Solve for the length of diagonal $BD$. $$8 = \frac{12d_2}{10} = \frac{6d_2}{5}$$ $$d_2 = \frac{8 \cdot 5}{6} = \frac{40}{6} = \frac{20}{3}$$ Therefore, the length of diagonal $BD$ is $\boxed{\frac{20}{3}}$ units, which corresponds to **Option 1**.