The line L has equation y = 5 - 2x - Edexcel - A-Level Maths Pure - Question 5 - 2006 - Paper 1

To determine the equation of the line perpendicular to L, we first need to find the slope of L.

The equation of line L can be rearranged as follows to find its slope:
$y = -2x + 5$

The slope of line L is $-2$. The slope of the line perpendicular to it will be the negative reciprocal:
m_{ ext{perpendicular}} = rac{1}{2}

Next, we use the point-slope form to find the equation of the perpendicular line that passes through point P(3, -1):
$y - y_1 = m(x - x_1)$ Where $(x_1, y_1) = (3, -1)$ and m = rac{1}{2}. Substituting these values gives:
y - (-1) = rac{1}{2}(x - 3)

Simplifying this:
y + 1 = rac{1}{2}x - rac{3}{2}
y = rac{1}{2}x - rac{5}{2}

To convert into the standard form $ext{αx + βy + c = 0}$, we rearrange:
rac{1}{2}x - y - rac{5}{2} = 0 Multiplying through by 2 to eliminate fractions, we have:
$x - 2y - 5 = 0$

Thus, the equation of the line perpendicular to L that passes through P is:
$x - 2y - 5 = 0$