Figure 2 shows a right angled triangle LMN - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 2

Since triangle LMN is a right-angled triangle and angle LMN = 90°, the slopes of the lines LM and MN should be negative reciprocals of each other.
From part (a), the slope of line LM is -\frac{3}{4}. Let the slope of line MN be m.
Thus, we have:
$m = -\frac{1}{\text{slope of LM}} = \frac{4}{3}$
The coordinates of point N are (16, p) and M is (7, -4). Therefore, the slope of line MN can be calculated as:
$\text{slope of MN} = \frac{p - (-4)}{16 - 7} = \frac{p + 4}{9}$
Setting the slopes equal gives:
$\frac{p + 4}{9} = \frac{4}{3}$
Cross-multiplying yields:
$3(p + 4) = 36$
Expanding gives:
$3p + 12 = 36$
Subtracting 12 from both sides:
$3p = 24$
Dividing by 3 gives:
$p = 8$

To find the y-coordinate of point K such that L, M, N, and K form a rectangle, we note that the diagonals of a rectangle bisect each other.
Let K be (x, y). The midpoint of diagonal LN must equal the midpoint of diagonal MK.
The coordinates of L are (-1, 2), M are (7, -4), and N are (16, 8):
Midpoint of LN:
$\left( \frac{-1 + 16}{2}, \frac{2 + 8}{2} \right) = \left( \frac{15}{2}, 5 \right)$
Midpoint of MK:
$\left( \frac{7 + x}{2}, \frac{-4 + y}{2} \right)$
Setting these midpoints equal gives two equations:
$\frac{7 + x}{2} = \frac{15}{2}$
Thus, solving for x gives:
$7 + x = 15 \Rightarrow x = 8$
And for the y-coordinates:
$\frac{-4 + y}{2} = 5 \Rightarrow -4 + y = 10 \Rightarrow y = 14$
Therefore, the y-coordinate of K is:
$y = 14$