Figure 2 shows the straight line l_1 with equation 4y = 5x + 12 - Edexcel - A-Level Maths Pure - Question 10 - 2018 - Paper 1

Since l_2 is parallel to l_1, it will have the same gradient. Therefore, we know that:

$m = \frac{5}{4}$

Now, we use the point E (12, 5) to find the y-intercept (c). Using the point-slope formula for the line:

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

Substituting in the values gives:

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

Expanding this, we find the equation of the line l_2:

$y = \frac{5}{4}x - \frac{5}{4} \cdot 12 + 5$

Calculating, we get:

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

To find point B, we set x = 0 in the equation of l_2:

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

Thus, the coordinates of point B are:

$(0, -10)$

To find point C, we set y = 0 in the equation of l_2:

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

Rearranging gives:

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

Multiplying by $\frac{4}{5}$ yields:

$x = \frac{4}{5} \cdot 10 = 8$

Thus, the coordinates of point C are:

$(8, 0)$

Since ABCD is a parallelogram, we can calculate its area using the formula:

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

Here, the base AC is length 8, derived from point C (8, 0) to A (0, 0). The height is the vertical distance from point B (0, -10) to the x-axis, which is 10. Thus, the area can be calculated as:

$\text{Area} = 8 \times 10 = 80$