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).