The line l_1, shown in Figure 2 has equation $2x + 3y = 26$ - Edexcel - A-Level Maths Pure - Question 11 - 2014 - Paper 1

To find the area of triangle OBC, we first need to identify the coordinates of points O, B, and C.

1. Coordinates of O: O is the origin, so O(0, 0).

2. Coordinates of B: To find point B where line l_1 intersects the y-axis, set $x = 0$ in the equation of line l_1:

$2(0) + 3y = 26 \Rightarrow 3y = 26 \Rightarrow y = \frac{26}{3}$

Thus, B(0, $\frac{26}{3}$).

3. Coordinates of C: To find point C, we set the equations of l_1 and l_2 equal:

$\frac{3}{2}x = -\frac{2}{3}x + \frac{26}{3}$

Multiplying through by 6 to eliminate fractions:

$9x = -4x + 52$ $13x = 52 \Rightarrow x = 4$

Plugging $x=4$ back into line l_2's equation to find y:

$y = \frac{3}{2}(4) = 6$

Thus, C(4, 6).

Now, we can find the area of triangle OBC using the formula:

$Area = \frac{1}{2} \times base \times height$

Using the base OB (along the y-axis) which is $\frac{26}{3}$ and the height OC (along the x-axis) which is 4:

$Area = \frac{1}{2} \times 4 \times \frac{26}{3} = \frac{52}{3}$

Thus, the area of triangle OBC is $\frac{52}{3}$, where $a = 52$ and $b = 3$.