Calculate the gradient of ${R_2}$ .

The point of intersection of ${R_1}$ and ${R_2}$ is $(4{\text{, }}k)$ .
Find
(i)     the value of $k$ ;
(ii)    the equation of ${R_2}$ .

$y = - 2x + 8$     (M1)

Note: Award (M1) for rearrangement of equation or for $- 2$ seen.

$m({\text{perp}}) = \frac{1}{2}$     (A1)     (C2)

[2 marks]

(i)     $2(4) + k - 8 = 0$     (M1)

Note: Award (M1) for evidence of substituting $x = 4$ into ${R_1}$ .

$k = 0$     (A1)     (C2)

(ii)   $y = \frac{1}{2}x + c$ (can be implied)     (M1)

Note: Award (M1) for substitution of $\frac{1}{2}$ into equation of the line.

$0 = \frac{1}{2}(4) + c$

$y = \frac{1}{2}x - 2$     (A1)(ft)     (C2)

Notes: Follow through from parts (a) and (b)(i). Accept equivalent forms for the equation of a line.

OR

$y - {y_1} = \frac{1}{2}(x - {x_1})$     (M1)

Note: Award (M1) for substitution of $\frac{1}{2}$ into equation of the line.

$y = \frac{1}{2}(x - 4)$     (A1)(ft)     (C2)

Notes: Follow through from parts (a) and (b)(i). Accept equivalent forms for the equation of a line.

[4 marks]

Parts a and bi of Question 13 appeared to be accessible to most candidates, but part bii was not well attempted. Many candidates did not show their working and lost method marks due to their incorrect answers.

Parts a and bi of Question 13 appeared to be accessible to most candidates, but part bii was not well attempted. Many candidates did not show their working and lost method marks due to their incorrect answers.