The line l_1, shown in Figure 2 has equation $2x + 3y = 26$ - Edexcel - A-Level Maths Pure - Question 10 - 2014 - Paper 1

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

• Point O (the origin) is (0, 0).

• Point B is where line l_1 intersects the y-axis: Substitute $x = 0$ into the equation of line l_1:

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

  So, point B is (0, $\frac{26}{3}$).

• Point C is found by substituting $l_2$ into $l_1$ to find their intersection:

  Substitute $y = \frac{3}{2}x$ into $2x + 3y = 26$:

  $2x + 3(\frac{3}{2}x) = 26$ $2x + \frac{9}{2}x = 26$ $\frac{13}{2}x = 26 \implies x = 4 \implies y = \frac{3}{2}(4) = 6$

So, point C is (4, 6).

Now, we can find area A of triangle OBC using the formula for the area of a triangle from coordinates:

$A = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substituting the coordinates O(0, 0), B(0, \frac{26}{3}), C(4, 6):

$A = \frac{1}{2} | 0(\frac{26}{3} - 6) + 0(6 - 0) + 4(0 - \frac{26}{3})|$ $A = \frac{1}{2} | 4(0 - \frac{26}{3})| \implies A = \frac{1}{2} \cdot 4 \cdot \frac{26}{3} = \frac{52}{3}$

Expressing the area in the form $\frac{a}{b}$, we have:

$A = \frac{52}{3}$ where a = 52 and b = 3.