For a clearer understanding of what is to be done, candidates were expected to sketch the diagram first.
From the diagram, gradient of BC =8+2/1-4 = -10/3 Gradient of AD = y-1/x+2. Since they are perpendicular, y-1/x+2 * -10/3 = -1 then 10y - 10 = 3x + 6 ; 10y – 3x - 16 = 0.
In part (b), length of BC = √ (8 + 2)2+ (1 - 4)2 = √109. The unit vector is obtained by:
(-3i + 10j) = √109 (-3i + 10j)
√109 109
The Chief Examiner reported that this question was not attempted by majority of the candidates. Some of those who attempted it displayed poor knowledge of the principle of perpendicularity of lines. Hence could not obtain the correct equation.
In part (b), although candidates displayed a good knowledge of the concept of unit vectors, many of them could not obtain the correct answer because they could not get the correct vector. 
