Find the coordinates of A.

Find the coordinates of B.

Show that the distance between A and B is 8.94 correct to 3 significant figures.

N lies on the line AB. The line CN is perpendicular to the line AB.

Find the gradient of CN.

N lies on the line AB. The line CN is perpendicular to the line AB.

Find the equation of CN.

N lies on the line AB. The line CN is perpendicular to the line AB.

Calculate the coordinates of N.

It is known that AC = 5 and BC = 8.06.

Calculate the size of angle ACB.

It is known that AC = 5 and BC = 8.06.

Calculate the area of triangle ACB.

A(0, 4)     Accept x = 0, y = 4     (A1)

[1 mark]

B(8, 0)     Accept x = 8, y = 0     (A1)(ft)

Note: Award (A0) if coordinates are reversed in (i) and (A1)(ft) in (ii).

[1 mark]

\({\text{AB}} = \sqrt {{8^2} + {4^2}} {\text{  }} = \sqrt {80} \)     (M1)

AB = 8.944     (A1)

= 8.94     (AG)

[2 marks]

y = –0.5x + 4     (M1)
Gradient AB = –0.5     (A1)

Note: Award (A2) if –0.5 seen.

OR

Gradient \({\text{AB}} = \frac{{(0 - 4)}}{{(8 - 0)}}\)     (M1)
\( = -\frac{1}{2}\)     (A1)

Note: Award (M1) for correct substitution in the gradient formula. Follow through from their answers to part (a).

Gradient CN = 2     (A1)(ft)(G2)

Note: Special case: Follow through for gradient CN from their gradient AB.

[3 marks]

CN: y = 2x + c

7 = 2(4) + c     (M1)

Note: Award (M1)for correct substitution in equation of a line.

y = 2x – 1     (A1)(ft)(G2)

Note: Accept alternative forms for the equation of a line including y – 7 = 2(x – 4) . Follow through from their gradient in (i).

Note: If c = –1 seen but final answer is not given, award (A1)(d).

[2 marks]

x + 2(2x – 1) = 8     or equivalent     (M1)

N(2, 3) (x = 2, y = 3)     (A1)(A1)(ft)(G3)

Note: Award (M1) for attempt to solve simultaneous equations or a sketch of the two lines with an indication of the point of intersection.

[3 marks]

Cosine rule: \(\cos ({\rm{A\hat CB)}} = \frac{{{5^2} + {{8.06}^2} - {{8.944}^2}}}{{2 \times 5 \times 8.06}}\)     (M1)(A1)

Note: Award (M1) for use of cosine rule with numbers from the problem substituted, (A1) for correct substitution.

\({\rm{A\hat CB  =  82.9^\circ }}\)     (A1)(G2)

Note: If alternative right-angled trigonometry method used award (M1) for use of trig ratio in both triangles, (A1) for correct substitution of their values in each ratio, (A1) for answer.

Note: Accept 82.8° with use of 8.94.

[3 marks]

Area \({\text{ACB}} = \frac{{5 \times 8.06\sin (82.9)}}{2}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted area formula, (A1) for correct substitution. Follow through from their angle in part (e).

OR

Area \({\text{ACB}} = \frac{{{\text{AB}} \times {\text{CN}}}}{2} = \frac{{8.94 \times \sqrt {{{(4 - 2)}^2} + {{(7 - 3)}^2}} }}{2}\)     (M1)(M1)(ft)

Note: Award (M1) substituted area formula with their values, (M1) for substituted distance formula. Follow through from
coordinates of N.

Area ACB = 20.0     (A1)(ft)(G2)

Note: Accept 20

[3 marks]

This question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question to the end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently from the rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.

This question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question to the end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently from the rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.

This question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question to the end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently from the rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.

This question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question to the end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently from the rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.

This question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question to the end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently from the rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.

This question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question to the end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently from the rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.

This question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question to the end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently from the rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.

This question had many correct solutions, but a large number of candidates were unable to follow the logical flow of the question to the end and many gave up. It should be pointed out to future candidates that parts (e) and (f) could be attempted independently from the rest and that care must be taken not to abandon hope too early in the longer questions of paper 2.