Figure 2 shows the straight line l₁ with equation 4y = 5x + 12 - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 1

Since l₂ is parallel to l₁, it has the same gradient, which is ( \frac{5}{4} ). The line passes through the point E (12, 5). We can use the point-slope form to find the equation:

$$y - y_1 = m(x - x_1)$$

Substituting the values, we have:

$$y - 5 = \frac{5}{4}(x - 12)$$

Expanding this gives:

$$y - 5 = \frac{5}{4}x - 15\ y = \frac{5}{4}x - 10$$

Thus, the equation of l₂ is ( y = \frac{5}{4}x - 10 ).

The point B is where l₂ cuts the y-axis, which occurs when x = 0:

$$y = \frac{5}{4}(0) - 10 = -10$$

Therefore, the coordinates of point B are (0, -10).

The point C is where l₂ cuts the x-axis, which occurs when y = 0:

$$0 = \frac{5}{4}x - 10$$

Rearranging gives us:

$$\frac{5}{4}x = 10$$

Multiplying through by ( \frac{4}{5} ) yields:

$$x = 8$$

Therefore, the coordinates of point C are (8, 0).

To find the area of parallelogram ABCD, we can use the formula for area:

$$\text{Area} = \text{base} \times \text{height}$$

The base (length of AB) can be determined as the distance from A (0, 3) to B (0, -10), which is:

$$3 - (-10) = 3 + 10 = 13$$

The height corresponds to the distance from C (8, 0) to the y-axis. Thus, it is 8 units.

Thus,

$$\text{Area} = 13 \times 8 = 104$$

Therefore, the area of ABCD is 104 square units.