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Date May 2009 Marks available 1 Reference code 09M.1.sl.TZ2.9
Level SL only Paper 1 Time zone TZ2
Command term Write down Question number 9 Adapted from N/A

Question

The coordinates of the vertices of a triangle ABC are A (4, 3), B (7, –3) and C (0.5, p).

Calculate the gradient of the line AB.

[2]
a.

Given that the line AC is perpendicular to the line AB

write down the gradient of the line AC.

[1]
b, i.

Given that the line AC is perpendicular to the line AB

find the value of p.

[3]
b, ii.

Markscheme

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


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

 

[2 marks]

a.

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

[1 mark]

b, i.

\(\frac{{p - 3}}{{0.5 - 4}} = \frac{1}{2}\)  (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]

b, ii.

Examiners report

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.

a.

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.

b, i.

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.

b, ii.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.1 » Gradient; intercepts.
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