Show that A lies on ${L_1}$.

Find the coordinates of M.

Find the length of AC.

Show that the equation of ${L_2}$ is $2y + x - 19 = 0$.

Find the coordinates of D.

Write down the length of MD correct to five significant figures.

Find the area of ABCD.

$2 \times 4 - 1 - 7 = 0$ (or equivalent)     (R1)

Note:     For (R1) accept substitution of $x = 1$ or $y = 4$ into the equation followed by a confirmation that $y = 4$ or $x = 1$.

(since the point satisfies the equation of the line,) A lies on ${L_1}$     (A1)

Note:     Do not award (A1)(R0).

[2 marks]

$\frac{{1 + 5}}{2}$ OR $\frac{{4 + 12}}{2}$ seen     (M1)

Note:     Award (M1) for at least one correct substitution into the midpoint formula.

$(3,{\text{ }}8)$    (A1)(G2)

Notes:     Accept $x = 3,{\text{ }}y = 8$.

Award (M1)(A0) for $\left( {\frac{{1 + 5}}{2},{\text{ }}\frac{{4 + 12}}{2}} \right)$.

Award (G1) for each correct coordinate seen without working.

[2 marks]

$\sqrt {{{(5 - 1)}^2} + {{(12 - 4)}^2}}$    (M1)

Note:     Award (M1) for a correct substitution into distance between two points formula.

$= 8.94{\text{ }}\left( {4\sqrt 5 ,{\text{ }}\sqrt {80} ,{\text{ }}8.94427 \ldots } \right)$    (A1)(G2)

[2 marks]

gradient of ${\text{AC}} = \frac{{12 - 4}}{{5 - 1}}$     (M1)

Note:     Award (M1) for correct substitution into gradient formula.

$= 2$    (A1)

Note:     Award (M1)(A1) for gradient of ${\text{AC}} = 2$ with or without working

gradient of the normal $=  - \frac{1}{2}$     (M1)

Note:     Award (M1) for the negative reciprocal of their gradient of AC.

$y - 8 =  - \frac{1}{2}(x - 3)$ OR $8 =  - \frac{1}{2}(3) + c$     (M1)

Note:     Award (M1) for substitution of their point and gradient into straight line formula. This (M1) can only be awarded where $- \frac{1}{2}$ (gradient) is correctly determined as the gradient of the normal to AC.

$2y - 16 =  - (x - 3)$ OR $- 2y + 16 = x - 3$ OR $2y =  - x + 19$     (A1)

Note:     Award (A1) for correctly removing fractions, but only if their equation is equivalent to the given equation.

$2y + x - 19 = 0$    (AG)

Note:     The conclusion $2y + x - 19 = 0$ must be seen for the (A1) to be awarded.

Where the candidate has shown the gradient of the normal to ${\text{AC}} =  - 0.5$, award (M1) for $2(8) + 3 - 19 = 0$ and (A1) for (therefore) $2y + x - 19 = 0$.

Simply substituting $(3,{\text{ }}8)$ into the equation of ${L_2}$ with no other prior working, earns no marks.

[5 marks]

$(6,{\text{ }}6.5)$    (A1)(A1)(G2)

Note:     Award (A1) for 6, (A1) for 6.5. Award a maximum of (A1)(A0) if answers are not given as a coordinate pair. Accept $x = 6,{\text{ }}y = 6.5$.

Award (M1)(A0) for an attempt to solve the two simultaneous equations $2y - x - 7 = 0$ and $2y + x - 19 = 0$ algebraically, leading to at least one incorrect or missing coordinate.

[2 marks]

3.3541     (A1)

Note:     Answer must be to 5 significant figures.

[1 mark]

$2 \times \frac{1}{2} \times \sqrt {80}  \times \frac{{\sqrt {45} }}{2}$     (M1)(M1)

Notes:     Award (M1) for correct substitution into area of triangle formula.

If their triangle is a quarter of the rhombus then award (M1) for multiplying their triangle by 4.

If their triangle is a half of the rhombus then award (M1) for multiplying their triangle by 2.

OR

$\frac{1}{2} \times \sqrt {80}  \times \sqrt {45}$    (M1)(M1)

Notes:     Award (M1) for doubling MD to get the diagonal BD, (M1) for correct substitution into the area of a rhombus formula.

Award (M1)(M1) for $\sqrt {80}  \times$ their (f).

$= 30$    (A1)(ft)(G3)

Notes:     Follow through from parts (c) and (f).

$8.94 \times 3.3541 = 29.9856 \ldots$

[3 marks]