The figure shows three points $A$, $C$ and $D$ whose coordinates are $(3, 5)$, $(4, -2)$ and $(7, 2)$ respectively.

1. Find the coordinates of the midpoint of $AC$.

Midpoint of $AC = \left( \frac{3 + 4}{2}, \frac{5 + (-2)}{2} \right) = \left( \frac{7}{2}, \frac{3}{2} \right)$

2. Find the gradient of $AC$ and the gradient of the perpendicular bisector.

Gradient of $AC = \frac{-2 - 5}{4 - 3} = -7$

Gradient of perpendicular bisector = $-\frac{1}{-7} = \frac{1}{7}$

3. Find the equation of the perpendicular bisector.

Using the point-slope form $y - y_1 = m(x - x_1)$:
$y - \frac{3}{2} = \frac{1}{7}\left(x - \frac{7}{2}\right)$
$y - \frac{3}{2} = \frac{1}{7}x - \frac{1}{2}$
$y = \frac{1}{7}x + 1$

1. Find the equation of the perpendicular bisector of $AC$ (result from part i)

Midpoint of $AC = \left(\frac{3+4}{2}, \frac{5+(-2)}{2}\right) = \left(\frac{7}{2}, \frac{3}{2}\right)$
Gradient of $AC = \frac{-2-5}{4-3} = -7$
Gradient of perpendicular bisector = $\frac{1}{7}$
Equation of perpendicular bisector: $y - \frac{3}{2} = \frac{1}{7}\left(x - \frac{7}{2}\right)$
$y = \frac{1}{7}x + 1$

2. Find the coordinates of $B$ by substituting $x = 0$ into the equation of the perpendicular bisector

Since $B$ lies on the $y$-axis, its $x$-coordinate is $0$.
Substitute $x = 0$ into $y = \frac{1}{7}x + 1$:
$y = \frac{1}{7}(0) + 1 = 1$
The coordinates of $B$ are $(0, 1)$.

1. State the coordinates of $B$ from part (ii)

From (ii), the coordinates of $B$ are $(0, 1)$.

2. Calculate the lengths of all four sides to show they are equal

$AB = \sqrt{(3 - 0)^2 + (5 - 1)^2} = \sqrt{9 + 16} = 5$
$BC = \sqrt{(4 - 0)^2 + (-2 - 1)^2} = \sqrt{16 + 9} = 5$
$CD = \sqrt{(7 - 4)^2 + (2 - (-2))^2} = \sqrt{9 + 16} = 5$
$AD = \sqrt{(7 - 3)^2 + (2 - 5)^2} = \sqrt{16 + 9} = 5$
Since $AB = BC = CD = AD = 5$, $ABCD$ is a rhombus.

3. Calculate the gradients of two adjacent sides to show they are perpendicular

Gradient of $AB$, $m_{AB} = \frac{5 - 1}{3 - 0} = \frac{4}{3}$
Gradient of $BC$, $m_{BC} = \frac{-2 - 1}{4 - 0} = -\frac{3}{4}$
$m_{AB} \times m_{BC} = \left(\frac{4}{3}\right) \times \left(-\frac{3}{4}\right) = -1$
Since the product of their gradients is $-1$, $AB \perp BC$, meaning $\angle ABC = 90^\circ$.

4. Conclude that the figure is a square

Since all four sides are equal in length and it contains a right angle, $ABCD$ is a square.