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 the line L₁:

$2y = 3x + k$

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

From this, the gradient (m) of L₁ is:

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

The gradient of L₂, being perpendicular to L₁, is:

$m_{L₂} = -\frac{2}{3}$

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

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

This equation simplifies to:

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

Thus:

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

Multiplying through by 3 to eliminate fractions:

$3y = -2x + 14$

Rearranging gives:

$2x + 3y - 14 = 0$

The line L₂ crosses the x-axis where y = 0. Setting y to 0 in the equation:

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

This gives:

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

Solving for x yields:

$x = 7$

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

Using the distance formula between points A (1, 4) and B (7, 0):

$AB = \sqrt{(7 - 1)^2 + (0 - 4)^2}$

Calculating gives:

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