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 2 - 2007 - Paper 2

To find the point of intersection ( P ) of the lines ( l_1 ) and ( l_2 ), we set their equations equal to each other:

1. Substitute ( y ) from ( l_1 ):
   [ 3x + 2 = -\frac{3}{2}x + 4 ]

2. Multiply through by 2 to eliminate the fraction:
   [ 6x + 4 = -3x + 8 ]
   [ 6x + 3x = 8 - 4 ]
   [ 9x = 4 ]
   [ x = \frac{4}{9} ]

3. Substitute ( x ) back into the equation of ( l_1 ) to find ( y ):
   [ y = 3\left(\frac{4}{9}\right) + 2 = \frac{12}{9} + \frac{18}{9} = \frac{30}{9} = \frac{10}{3} ]

Thus, the coordinates of ( P ) are ( P\left(\frac{4}{9}, \frac{10}{3}\right) ).

To find the area of triangle ( ABP ), we first need to find the points ( A ) and ( B ).

1. For point ( A ), substitute ( y = 1 ) into the equation of ( l_1 ):
   [ 1 = 3x + 2 ]
   [ 3x = 1 - 2 ]
   [ 3x = -1 ]
   [ x = -\frac{1}{3} ]
   [ A\left(-\frac{1}{3}, 1\right) ]

2. For point ( B ), substitute ( y = 1 ) into the equation of ( l_2 ):
   [ 3x + 2(1) - 8 = 0 ]
   [ 3x + 2 - 8 = 0 ]
   [ 3x = 6 ]
   [ x = 2 ]
   [ B(2, 1) ]

3. Now, we can calculate the area of triangle ( ABP ) using the formula for the area given by vertices ( (x_1, y_1), (x_2, y_2), (x_3, y_3) ):
   [ \text{Area} = \frac{1}{2}\left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| ]

   Substituting the coordinates:
   [ \text{Area} = \frac{1}{2} \left| -\frac{1}{3}\left(1 - \frac{10}{3}\right) + 2\left(\frac{10}{3} - 1\right) + \frac{4}{9}\left(1 - 1\right) \right| ] [ = \frac{1}{2} \left| -\frac{1}{3}\left(-\frac{7}{3}\right) + 2\left(\frac{7}{3}\right) \right| ]
   [ = \frac{1}{2}\left| \frac{7}{9} + \frac{14}{3} \right| = \frac{1}{2}\left| \frac{7}{9} + \frac{42}{9} \right| = \frac{1}{2}\left| \frac{49}{9} \right| = \frac{49}{18} ]

Thus, the area of triangle ( ABP ) is ( \frac{49}{18} ).