Calculate the gradient of the line AB.

Given that the line AC is perpendicular to the line AB

write down the gradient of the line AC.

Given that the line AC is perpendicular to the line AB

find the value of p.

\(m({\rm{AB}}) = \frac{{ - 3 - 3}}{{7 - 4}} = - 2\)     (M1)(A1)     (C2)

Note: Award (M1) for attempt to substitute into correct gradient formula.

[2 marks]

\(m({\rm{AC}}) = \frac{1}{2}\)     (A1)(ft)

[1 mark]

\(\frac{{p - 3}}{{0.5 - 4}} = \frac{1}{2}\)  (or equivalent method)     (M1)(A1)(ft)

Note: Award (M1) for equating gradient to \(\frac{1}{2}\). (A1) for correct substitution.

\(p = 1.25\)     (A1)(ft)     (C4)

[3 marks]

While parts (a) and (b)(i) were attempted with some success, few candidates made progress in (b)(ii). Some candidates used the coordinates of point B rather than C and others could not find the unknown value p as they did not realise they had to equate their substituted formula for the gradient to the answer to part (b)(i). A large number of candidates did not attempt this part of the question.

While parts (a) and (b)(i) were attempted with some success, few candidates made progress in (b)(ii). Some candidates used the coordinates of point B rather than C and others could not find the unknown value p as they did not realise they had to equate their substituted formula for the gradient to the answer to part (b)(i). A large number of candidates did not attempt this part of the question.

While parts (a) and (b)(i) were attempted with some success, few candidates made progressin (b)(ii). Some candidates used the coordinates of point B rather than C and others could not find the unknown value p as they did not realise they had to equate their substituted formula for the gradient to the answer to part (b)(i). A large number of candidates did notattempt this part of the question.