Figure 2 shows the straight line $l_1$ with equation $4y = 5x + 12$ (a) State the gradient of $l_1$ - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 1

Since $l_1$ is parallel to the original line, it shares the same gradient of rac{5}{4}. We need to find the equation of the line that passes through the point $E(12, 5)$. Using the point-slope form of a line:

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

Substituting the values:

y - 5 = rac{5}{4}(x - 12)

Expanding this, we get:

y - 5 = rac{5}{4}x - 15

Adding 5 to both sides gives:

y = rac{5}{4}x - 10

Thus, the equation of $l_1$ is:

y = rac{5}{4}x - 10

To find the y-intercept (point $B$), we set $x = 0$ in the equation of $l_1$:

y = rac{5}{4}(0) - 10 = -10

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

To find the x-intercept (point $C$), we set $y = 0$ in the equation of $l_1$:

0 = rac{5}{4}x - 10

Solving for $x$, we have:

rac{5}{4}x = 10 \implies x = 8

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

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

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

The base $AB$ has a length of:

$|y_B - y_A| = |-10 - 0| = 10$

The height ($l_1$) can be found using the x-coordinate difference between $C$ and $A$:

$|x_C - x_A| = |8 - 0| = 8$

Therefore, the area is given by:

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

Thus, the area of the parallelogram $ABCD$ is 80.