The straight line L1 has equation y = 3x - 4 - Edexcel - GCSE Maths - Question 16 - 2020 - Paper 1

Since line L2 is perpendicular to line L1, we can find its gradient (slope) by using the negative reciprocal of $m_1$. Therefore, the gradient of L2, denoted as $m_2$, is:

$m_2 = -\frac{1}{m_1} = -\frac{1}{3}$

We will use the point-slope form of a line equation, which is given as:

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

Substituting the point (9, 5) and the gradient $m_2$ into the equation:

$y - 5 = -\frac{1}{3} (x - 9)$

Expanding and rearranging gives:

$y - 5 = -\frac{1}{3}x + 3$

Now, adding 5 to both sides results in:

$y = -\frac{1}{3}x + 8$

Thus, the equation of line L2 is:

$y = -\frac{1}{3}x + 8$