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]