This problem could be solved using the vector method i.e
->
{ 2 – 2} { 0 } -> { 7 – x}
AB = { 2 – 6} = {- 4 } ; DC = { 3 – y} . This implies that 7 – x = 0 or x = 7
and 3 – y = - 4 » y = 7, thus D ( 7, 7 ) .
It could also be solved using the gradient method which was the most popular method i.e Gradient of DA
= y – 6 , Gradient of CB = 3 – 2 = 1
x - 2 7 – 2 5
Since DA is parallel to CB, then y – 6 = 1 » y – 6 = 1 and x – 2 = 5
x – 2 5
» D = (x, y) = ( 7, 7).
Although many candidates did it correctly, a good number of them could not make it because they could not link up the two ideas of parallelism and equality. One common error which was also noticed was that some candidates made the mistake of stating that gradient of a line is Δx as against Δy. Δy Δx
