The points P(0, 2) and Q(3, 7) lie on the line l_1, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 11 - 2016 - Paper 1

To find the coordinates of point R, we set y = 0 in the equation of line l_2:

1. **Substitute y = 0 into the equation: ** 3x - 44 = 0 \ 3x = 44 \ x = \frac{44}{3}$$
2. Thus, the coordinates of R are: $R(\frac{44}{3}, 0)$

To find the area of the quadrilateral ORQP, we can divide it into two triangles: OQP and ORP.

1. Coordinates of points:

   • O(0, 0)
   • P(0, 2)
   • Q(3, 7)
   • R(\frac{44}{3}, 0)

2. Area of triangle OQP: Using the formula for the area of a triangle: $A = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |$ where (x_1, y_1), (x_2, y_2), and (x_3, y_3) are the coordinates of points O, Q, and P: $A_{OQP} = \frac{1}{2} | 0(7 - 2) + 3(2 - 0) + 0(0 - 7) | = \frac{1}{2} | 0 + 6 + 0 | = 3$

3. Area of triangle ORP: Using the same formula: $A_{ORP} = \frac{1}{2} | 0(2 - 0) + \frac{44}{3}(0 - 2) + 0(0 - 2) | = \frac{1}{2} | 0 - \frac{88}{3} + 0 | = \frac{44}{3}$

4. Total Area: $A_{total} = A_{OQP} + A_{ORP} = 3 + \frac{44}{3} = \frac{9}{3} + \frac{44}{3} = \frac{53}{3}$ Therefore, the exact area of quadrilateral ORQP is: $\frac{53}{3}$