The line $l_1$ has equation $2x - 3y + 12 = 0$ - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 1

To find the coordinates of point B where the line crosses the y-axis, we set $x = 0$ in the equation:

$y = \frac{2}{3}(0) + 4 = 4$

Thus, the coordinates of B are (0, 4).

Since $l_1$ has a gradient of rac{2}{3}, the gradient of line $l_2$, which is perpendicular to $l_1$, can be found using the fact that the product of the slopes of perpendicular lines is $-1$:

$m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}$

Now using the point B(0, 4), we employ the point-slope form of the line equation:

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

Substituting gives:

$y - 4 = -\frac{3}{2}(x - 0) \implies y = -\frac{3}{2}x + 4$

Thus, the equation of $l_2$ is $y = -\frac{3}{2}x + 4$.

To find where line $l_2$ crosses the x-axis (point C), we set $y = 0$ in the equation of $l_2$:

$0 = -\frac{3}{2}x + 4$

Solving for $x$ gives:

$\frac{3}{2}x = 4 \implies x = \frac{4}{\frac{3}{2}} = \frac{4 \cdot 2}{3} = \frac{8}{3}$

Therefore, the coordinates of C are $\left( \frac{8}{3}, 0 \right)$.

To find the area of triangle ABC, we can use the formula:

$\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}$

Taking AC as base with length $\frac{8}{3}$ and the height from B (which is the y-coordinate of point B):

$\text{Area} = \frac{1}{2} \cdot \frac{8}{3} \cdot 4 = \frac{32}{6} = \frac{16}{3}$

Thus, the area of triangle ABC is $\frac{16}{3}$.