The line L₁ has equation 2y−3x−k=0, where k is a constant - Edexcel - A-Level Maths Pure - Question 11 - 2011 - Paper 2

To determine the gradient of L₁, we rewrite the equation in slope-intercept form (y = mx + c).

Starting with:

$2y - 3x - 5 = 0$

We rearrange to get:

\text{or } y = \frac{3}{2}x + \frac{5}{2}$$ Thus, the gradient (m) of L₁ is \( \frac{3}{2} \).

Since L₂ is perpendicular to L₁, its gradient is the negative reciprocal of ( \frac{3}{2} ):

$m_{L₂} = -\frac{2}{3}$

Using the point-slope form of the line equation, where we substitute for point A(1, 4):

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

Simplifying, we get:

\implies y = -\frac{2}{3}x + \frac{14}{3}$$ Multiplying through by 3 to eliminate the fraction results in: $$3y + 2x - 14 = 0$$

The line L₂ crosses the x-axis where y = 0. Thus, substituting y = 0 into the equation of L₂:

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

Multiplying through by 3 gives:

\implies 2x = 14 \\ \implies x = 7$$ Hence, the coordinates of B are (7, 0).

To find the length of segment AB, we use the distance formula:

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

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

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