Solution to this question by accurate drawing will not be accepted. The diagram shows a kite $ABCD$ with $AB = AD$ and $CB = CD$. The diagonals intersect at $M$. It is given that the coordinates of $A$ and $B$ are $(0, 6)$ and $(-5, 1)$ respectively and the equation of $AC$ is $2x + y = 6$. Find Given further that the area of the triangle $ABD$ is $\frac{1}{3}$ of the area of the triangle $CBD$,

1. Find the gradient of the diagonal AC

The equation of $AC$ is $2x + y = 6$, which can be written as $y = -2x + 6$.
The gradient of $AC$, $m_{AC} = -2$.

2. Find the gradient of the diagonal BD

In a kite, the diagonals intersect at right angles. Therefore, $BD$ is perpendicular to $AC$.
$m_{BD} = -\frac{1}{m_{AC}} = -\frac{1}{-2} = \frac{1}{2}$.

3. Find the equation of BD

The line $BD$ passes through $B(-5, 1)$ with gradient $\frac{1}{2}$.
$y - 1 = \frac{1}{2}(x - (-5))$
$y - 1 = \frac{1}{2}x + \frac{5}{2}$
$y = \frac{1}{2}x + \frac{7}{2}$

1. Find the equation of $BD$ (from part (i))

The gradient of $AC$ ($2x + y = 6 \implies y = -2x + 6$) is $-2$.
Since the diagonals of a kite are perpendicular, the gradient of $BD$ is $\frac{1}{2}$.
Equation of $BD$ passing through $B(-5, 1)$:
$y - 1 = \frac{1}{2}(x - (-5))$
$y = \frac{1}{2}x + \frac{7}{2}$

2. Solve simultaneously to find the coordinates of $M$

Substitute $y = -2x + 6$ into the equation of $BD$:
$-2x + 6 = \frac{1}{2}x + \frac{7}{2}$
$\frac{5}{2}x = \frac{5}{2}$
$x = 1$
Substitute $x = 1$ into $y = -2x + 6$:
$y = -2(1) + 6 = 4$
Therefore, $M$ is $(1, 4)$.

3. Use the properties of a kite to relate $M$ to $B$ and $D$

In kite $ABCD$ with $AB = AD$ and $CB = CD$, the diagonal $AC$ is the perpendicular bisector of $BD$.
Thus, $M$ is the midpoint of $BD$.

4. Find the coordinates of $D$ using the midpoint formula

Let the coordinates of $D$ be $(x, y)$.
$\left( \frac{-5 + x}{2}, \frac{1 + y}{2} \right) = (1, 4)$
$\frac{-5 + x}{2} = 1 \implies -5 + x = 2 \implies x = 7$
$\frac{1 + y}{2} = 4 \implies 1 + y = 8 \implies y = 7$
Therefore, $D$ is $(7, 7)$.

1. Relate the areas of the triangles to the lengths of the segments on the diagonal $AC$.

The area of $\triangle ABD = \frac{1}{2} \times BD \times AM$ and the area of $\triangle CBD = \frac{1}{2} \times BD \times CM$.
Since Area of $\triangle ABD = \frac{1}{3} \times$ Area of $\triangle CBD$, we have:
$\frac{1}{2} \times BD \times AM = \frac{1}{3} \left( \frac{1}{2} \times BD \times CM \right)$
$AM = \frac{1}{3} CM \implies CM = 3AM$

2. Find the coordinates of $M$ from earlier parts.

From earlier parts, the equation of $BD$ is $y = \frac{1}{2}x + \frac{7}{2}$.
Solving simultaneously with $AC$ ($2x + y = 6$):
$2x + \frac{1}{2}x + \frac{7}{2} = 6 \implies \frac{5}{2}x = \frac{5}{2} \implies x = 1$
$y = 4$
So $M = (1, 4)$.

3. Use the ratio of lengths to find the coordinates of $C$.

Since $A, M, C$ are collinear in that order, the vector $\vec{MC} = 3\vec{AM}$.
$\vec{AM} = \begin{pmatrix} 1 - 0 \\ 4 - 6 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$
$\vec{MC} = 3 \begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \end{pmatrix}$
$\vec{OC} = \vec{OM} + \vec{MC} = \begin{pmatrix} 1 \\ 4 \end{pmatrix} + \begin{pmatrix} 3 \\ -6 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$
The coordinates of $C$ are $(4, -2)$.

1. Recall the coordinates of the vertices of the kite from earlier parts

From earlier parts, the coordinates of the vertices are $A(0, 6)$, $B(-5, 1)$, $C(4, -2)$, and $D(7, 7)$.

2. Apply the shoelace formula to find the area of the kite $ABCD$

$\begin{aligned} \text{Area} &= \frac{1}{2} \left| \begin{matrix} 0 & -5 & 4 & 7 & 0 \\ 6 & 1 & -2 & 7 & 6 \end{matrix} \right| \\ &= \frac{1}{2} | (0(1) + (-5)(-2) + 4(7) + 7(6)) - (6(-5) + 1(4) + (-2)(7) + 7(0)) | \\ &= \frac{1}{2} | (0 + 10 + 28 + 42) - (-30 + 4 - 14 + 0) | \\ &= \frac{1}{2} | 80 - (-40) | \\ &= \frac{1}{2} (120) \\ &= 60 \end{aligned}$