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 need the coordinates of points $B$ and $C$.

First, we find point $B$, where line $l_1$ intersects the y-axis. Setting $x = 0$ in the equation of line $l_1$:

$2(0) + 3y = 26 \Rightarrow 3y = 26 \Rightarrow y = \frac{26}{3}$ Thus, point $B$ is $(0, \frac{26}{3})$.

Next, we find point $C$, the intersection of lines $l_1$ and $l_2$. We set the equations equal to solve for $x$:

(\frac{3}{2}x = -\frac{2}{3}x + \frac{26}{3}) Multiply through by 6 to eliminate fractions:

$9x = -4x + 52 \Rightarrow 13x = 52 \Rightarrow x = 4$ Now substitute $x = 4$ back into the equation of line $l_2$ to get $y$:

$y = \frac{3}{2}(4) = 6$ Therefore, point $C$ is $(4, 6)$.

The vertices of triangle $OBC$ are now identified as:

• $O(0, 0)$
• $B(0, \frac{26}{3})$
• $C(4, 6)$

Using the formula for the area of a triangle given by vertices at $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$ Substituting the coordinates:

$\text{Area} = \frac{1}{2} \left| 0\left(\frac{26}{3} - 6\right) + 0\left(6 - 0\right) + 4\left(0 - \frac{26}{3}\right) \right|$ This simplifies to:

$\text{Area} = \frac{1}{2} \left| 4 \left( -\frac{26}{3} \right) \right| = \frac{1}{2} \left( \frac{104}{3} \right) = \frac{52}{3}$ Consequently, the area of triangle $OBC$ is:

$\text{Area} = \frac{52}{3}$