The line L1 has equation 4x + 2y - 3 = 0 (a) Find the gradient of L1 - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 2

Given that the line L2 is perpendicular to L1, the gradient of L2 can be found using the relationship of perpendicular gradients:

If the gradient of line L1 is m1, then the gradient of line L2 (m2) is:

$m_2 = -\frac{1}{m_1} = -\frac{1}{-2} = \frac{1}{2}$

Now we have the gradient for L2, which is: $m = \frac{1}{2}$

The equation of line L2 in point-slope form is:

$y - y_1 = m(x - x_1)$

Substituting the point (2, 5):

$y - 5 = \frac{1}{2}(x - 2)$

Distributing:

$y - 5 = \frac{1}{2}x - 1$

Adding 5 to both sides gives:

$y = \frac{1}{2}x + 4$

Thus, the equation of L2 in the form y = mx + c is: $y = \frac{1}{2}x + 4$