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 3 - 2007 - Paper 1

To find the point of intersection P of lines $l_1$ and $l_2$, we set the equations equal:

$3x + 2 = -\frac{3}{2}x + 4$

Solving for $x$:

$3x + \frac{3}{2}x = 4 - 2$ $\left(3 + \frac{3}{2}\right)x = 2$ $\frac{9}{2}x = 2$ $x = \frac{4}{9}$

Now substituting $x = \frac{4}{9}$ back into the equation of line $l_1$ 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 point P are $P\left(\frac{4}{9}, \frac{10}{3}\right)$.

To find the area of triangle ABP, we first determine the coordinates of points A and B by substituting $y = 1$ into the equations of lines $l_1$ and $l_2$ respectively.

For point A on line $l_1$:

$1 = 3x + 2$ $3x = 1 - 2$ $x = -\frac{1}{3}$

Therefore, $A\left(-\frac{1}{3}, 1\right)$.

For point B on line $l_2$:

$3x + 2(1) - 8 = 0$ $3x - 6 = 0$ $x = 2$

Thus, $B(2, 1)$.

Now, we have points A, B, and P:

• A: $\left(-\frac{1}{3}, 1\right)$
• B: $(2, 1)$
• P: $\left(\frac{4}{9}, \frac{10}{3}\right)$

Using the formula for the area of a triangle:

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

$Area = \frac{1}{2}\left| -\frac{1}{3}(1-\frac{10}{3}) + 2(\frac{10}{3}-1) + \frac{4}{9}(1-1) \right|$

Simplifying:

$= \frac{1}{2}\left| -\frac{1}{3}(\frac{-7}{3}) + 2(\frac{10}{3}-\frac{3}{3}) \right|$

$= \frac{1}{2}\left| \frac{7}{9} + 2(\frac{7}{3}) \right|$ $= \frac{1}{2}\left| \frac{7}{9} + \frac{14}{3} \right|$ $= \frac{1}{2}\left| \frac{7}{9} + \frac{42}{9} \right|$ $= \frac{1}{2}\left| \frac{49}{9} \right|$

$= \frac{49}{18}$

Therefore, the area of triangle ABP is $\frac{49}{18}$.