The line L1 has equation 2y - 3x - k = 0, where k is a constant - Edexcel - A-Level Maths Pure - Question 10 - 2011 - Paper 2

Rearranging the equation of line L1:

$2y = 3x + k$

Divide by 2 to express it in slope-intercept form:

$y = \frac{3}{2}x + \frac{k}{2}$

Thus, the gradient (m) of L1 is:

$m = \frac{3}{2}$

The gradient of line L2, which is perpendicular to L1, can be calculated as:

$m_{L2} = -\frac{1}{m_{L1}} = -\frac{2}{3}$

Using point-slope form with point A (1, 4):

$y - 4 = -\frac{2}{3}(x - 1)$

Expanding this:

$y - 4 = -\frac{2}{3}x + \frac{2}{3} \\ y = -\frac{2}{3}x + \frac{14}{3}$

To convert to the form ax + by + c = 0:

$2x + 3y - 14 = 0$

To find the coordinates of point B, set y = 0 in the equation of line L2:

$0 = -\frac{2}{3}x + \frac{14}{3}$

This simplifies to:

$\frac{2}{3}x = \frac{14}{3} \\ x = 7$

Thus, the coordinates of B are (7, 0).

Using the distance formula, the length of AB is calculated as follows:

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substituting the coordinates A (1, 4) and B (7, 0):

$AB = \sqrt{(7 - 1)^2 + (0 - 4)^2} \\ = \sqrt{(6)^2 + (-4)^2} \\ = \sqrt{36 + 16} \\ = \sqrt{52} \\ = 2\sqrt{13}$