The equation of the line BC is \(y = 4\).

Write down the coordinates of point C.

The \(x\)-coordinate of point B is \(a\).

(i)     Write down the coordinates of point B;

(ii)     Write down the gradient of the line OB.

Point A lies on the \(x\)-axis and the line AB is perpendicular to line OB.

(i)     Write down the gradient of line AB.

(ii)     Show that the equation of the line AB is \(4y + ax - {a^2} - 16 = 0\).

The area of triangle AOB is three times the area of triangle OBC.

Find an expression, in terms of a, for

(i)     the area of triangle OBC;

(ii)     the x-coordinate of point A.

Calculate the value of \(a\).

\((0,{\text{ }}4)\)     (A1)

Notes: Accept \(x = 0,{\text{ }}y = 4\).

(i)     \((a,{\text{ }}4)\)     (A1)(ft)

Notes: Follow through from part (a).

(ii)     \(\frac{4}{a}\)     (A1)(ft)

Note: Follow through from part (b)(i).

(i)     \( - \frac{a}{4}\)     (A1)(ft)

Note: Follow through from part (b)(ii).

(ii)     \(y =  - \frac{a}{4}x + c\)     (M1)

Note: Award (M1) for substitution of their gradient from part (c)(i) in the equation.

\(4 =  - \frac{a}{4} \times a + c\)

\(c = \frac{1}{4} \times {a^2} + 4\)

\(y =  - \frac{a}{4}x + \frac{1}{4}{a^2} + 4\)     (A1)

OR

\(y - 4 =  - \frac{a}{4}(x - a)\)     (M1)

Note: Award (M1) for substitution of their gradient from part (c)(i) in the equation.

\(y =  - \frac{{ax}}{4} + \frac{{{a^2}}}{4} + 4\)     (A1)

\(4y =  - ax + {a^2} + 16\)

\(4y + ax - {a^2} - 16 = 0\)     (AG)

Note: Both the simplified and the not simplified equations must be seen for the final (A1) to be awarded.

(i)     \(2a\)     (A1)

(ii)     \(\frac{{4x}}{2} = 3 \times 2a\)     (M1)

Note: Award (M1) for correct equation.

\(x = 3a\)     (A1)(ft)

Note: Follow through from part (d)(i).

OR

\(0 - 4 =  - \frac{a}{4}(x - a)\)     (M1)

Note: Award (M1) for correct substitution of their gradient and the coordinates of their point into the equation of a line.

\(\frac{{16}}{a} = x - a\)

\(x = a + \frac{{16}}{a}\)     (A1)(ft)

Note: Follow through from parts (b)(i) and (c)(i).

OR

\(4 \times 0 + ax - {a^2} - 16 = 0\)     (M1)

Note: Award (M1) for correct substitution of the coordinates of \({\text{A}}(x,{\text{ }}0)\) into the equation of line AB.

\(ax - {a^2} - 16 = 0\)

\(x = a + \frac{{16}}{a}\;\;\;\)OR\(\;\;\;x = \frac{{({a^2} + 16)}}{a}\)     (A1)(G1)

\(4(0) + a(3a) - {a^2} - 16 = 0\)     (M1)

Note: Award (M1) for correct substitution of their \(3a\) from part (d)(ii) into the equation of line AB.

OR

\(\frac{1}{2}\left( {a + \frac{{16}}{a}} \right) \times 4 = 3\left( {\frac{{4a}}{2}} \right)\)     (M1)

Note: Award (M1) for area of triangle AOB (with their substituted \(a + \frac{{16}}{a}\) and 4) equated to three times their area of triangle AOB.

\(a = 2.83\;\;\;\left( {2.82842...,{\text{ }}2\sqrt 2 ,{\text{ }}\sqrt 8 } \right)\)     (A1)(ft)(G1)

Note: Follow through from parts (d)(i) and (d)(ii).