The line l_1 passes through the point (9, -4) and has gradient \frac{1}{3}; (a) Find an equation for l_1 in the form ax + by + c = 0, where a, b and c are integers - Edexcel - A-Level Maths Pure - Question 9 - 2005 - Paper 1

For line l_2, which passes through the origin O and has a gradient of -2, its equation is:

$y = -2x$

Now, we need to find the intersection point P of l_1 and l_2 by solving the equations:

1. From l_1: (y = \frac{1}{3}x - 7)
2. From l_2: (y = -2x)

Setting these two equations equal to each other:

$\frac{1}{3}x - 7 = -2x$

Multiplying through by 3 to eliminate the fraction:

$x - 21 = -6x$

Combining like terms gives:

$7x = 21 \Rightarrow x = 3$

Substituting (x = 3) back into the equation of l_2:

$y = -2(3) = -6$

Thus, the coordinates of P are (3, -6).

To find the area of triangle OCP, where O is (0, 0), C is (0, -7), and P is (3, -6), we can use the formula for the area of a triangle given by vertices:

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

Substituting in the coordinates:

• ( O(0, 0) ) ( C(0, -7) ) ( P(3, -6) )

This gives:

$\text{Area} = \frac{1}{2} | 0(-7 - (-6)) + 0(-6 - 0) + 3(0 - (-7))|$

Simplifying further provides:

$\text{Area} = \frac{1}{2} |0 + 0 + 21| = \frac{21}{2}$

Thus, the exact area of triangle OCP is ( \frac{21}{2} ).