The line l_1 has equation y = 3x + 2 and the line l_2 has equation 3x + 2y - 8 = 0 - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 1

To find the intersection point P of the lines l_1 and l_2, we set their equations equal:

1. From line l_1:
   $y = 3x + 2$
2. From line l_2 (rearranged):
   $y = -\frac{3}{2}x + 4$

Setting the two equations equal gives:

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

Combining like terms:
$3x + \frac{3}{2}x = 4 - 2$
Multiply by 2 to eliminate the fraction:

$6x + 3x = 4$
$9x = 2$
Thus,
$x = \frac{2}{9}$

Plugging $x$ back into line l_1 to find y:

$y = 3\left(\frac{2}{9}\right) + 2 = \frac{6}{9} + \frac{18}{9} = \frac{24}{9} = \frac{8}{3}$

Therefore, the coordinates of P are:
$P\left(\frac{2}{9}, \frac{8}{3}\right)$.

To find the points A and B where lines l_1 and l_2 intersect y = 1, we set y = 1 in both equations:

For line l_1: $1 = 3x + 2$ Rearranging: $3x = -1 \\ x = -\frac{1}{3}$ Thus, A is at: $A\left(-\frac{1}{3}, 1\right)$

For line l_2: $1 = -\frac{3}{2}x + 4$ Rearranging: $-\frac{3}{2}x = 1 - 4 \\ -\frac{3}{2}x = -3 \\ x = 2$ Thus, B is at: $B\left(2, 1\right)$.

Now we have the coordinates for A, B, and P. The area of triangle ABP can be found using the formula: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2|$ Where A = (x_1, y_1), B = (x_2, y_2), and P = (x_3, y_3):
A = $(-\frac{1}{3}, 1)$, B = $(2, 1)$, P = $(\frac{2}{9}, \frac{8}{3})$.

Substituting in: $\text{Area} = \frac{1}{2} \left| -\frac{1}{3} \left( 1 - \frac{8}{3} \right) + 2 \left( \frac{8}{3} - 1 \right) + \frac{2}{9} \left( 1 - 1 \right) \right|$

Calculating the area: $= \frac{1}{2} \left| -\frac{1}{3} \left( -\frac{5}{3} \right) + 2 \left( \frac{5}{3} \right) \right| = \frac{1}{2} \left| \frac{5}{9} + \frac{10}{3} \right| = \frac{1}{2} \left| \frac{5 + 30}{9} \right| = \frac{1}{2} \left| \frac{35}{9} \right| = \frac{35}{18}$.

Thus, the area of triangle ABP is $\frac{35}{18}$.