The diagram shows a kite $ABCD$ in which the line $AC$ passes through the origin $O$ and has a gradient of 1.5.

1. Set up the equation for the gradient of line AC using the coordinates of C and the origin.

The line $AC$ passes through the origin $O(0,0)$ and $C(-2, k)$.
The gradient of $AC$ is given as $1.5$.
Gradient $m = \frac{y_2 - y_1}{x_2 - x_1}$
$1.5 = \frac{k - 0}{-2 - 0}$

2. Solve for k.

$1.5 = \frac{k}{-2}$
$k = 1.5 \times (-2)$
$k = -3$

1. Determine the gradient of the line $BD$

The diagonals of a kite are perpendicular to each other.
Given the gradient of $AC$ is $1.5$ (or $\frac{3}{2}$),
Gradient of $BD = -\frac{1}{1.5} = -\frac{2}{3}$

2. Find the equation of the line $BD$

Using the point $B(1, -2)$ and gradient $m = -\frac{2}{3}$:
$y - y_1 = m(x - x_1)$
$y - (-2) = -\frac{2}{3}(x - 1)$
$y + 2 = -\frac{2}{3}x + \frac{2}{3}$
$y = -\frac{2}{3}x - \frac{4}{3}$

1. Find the equations of lines $AC$ and $BD$ (from earlier parts).

The line $AC$ passes through the origin with a gradient of $1.5$, so its equation is $y = \frac{3}{2}x$.
Since $ABCD$ is a kite, its diagonals are perpendicular. The gradient of $BD$ is $-\frac{1}{1.5} = -\frac{2}{3}$.
Line $BD$ passes through $B(1, -2)$. Its equation is:
$y - (-2) = -\frac{2}{3}(x - 1)$
$y = -\frac{2}{3}x - \frac{4}{3}$

2. Solve the equations of $AC$ and $BD$ simultaneously to find the $x$-coordinate of $M$.

Equating the two expressions for $y$:
$\frac{3}{2}x = -\frac{2}{3}x - \frac{4}{3}$
Multiply the entire equation by $6$ to clear denominators:
$9x = -4x - 8$
$13x = -8$
$x = -\frac{8}{13}$

3. Find the corresponding $y$-coordinate of $M$.

Substitute $x = -\frac{8}{13}$ into the equation of $AC$:
$y = \frac{3}{2}\left(-\frac{8}{13}\right)$
$y = -\frac{12}{13}$
Thus, the coordinates of $M$ are $\left(-\frac{8}{13}, -\frac{12}{13}\right)$.

1. Identify the relationship between $MD$ and $MB$ using the properties of a kite.

In kite $ABCD$, the diagonal $AC$ is the axis of symmetry (since $AB \neq CB$), so it perpendicularly bisects $BD$. Therefore, $M$ is the midpoint of $BD$, which means $MD = MB$, and thus $MD^2 = MB^2$.

2. Calculate $MB^2$.

$MB^2 = \left(1 - \left(-\frac{8}{13}\right)\right)^2 + \left(-2 - \left(-\frac{12}{13}\right)\right)^2$
$MB^2 = \left(\frac{21}{13}\right)^2 + \left(-\frac{14}{13}\right)^2$
$MB^2 = \frac{441}{169} + \frac{196}{169} = \frac{637}{169} = \frac{49}{13}$

3. Express $MD^2$ in terms of $a$ and $b$ and equate to $MB^2$ to prove the statement.

$MD^2 = \left(a - \left(-\frac{8}{13}\right)\right)^2 + \left(b - \left(-\frac{12}{13}\right)\right)^2$
$\left(a + \frac{8}{13}\right)^2 + \left(b + \frac{12}{13}\right)^2 = \frac{49}{13}$
Multiplying both sides by 13:
$13\left(a + \frac{8}{13}\right)^2 + 13\left(b + \frac{12}{13}\right)^2 = 49$ (Shown)