Solutions to this equation by accurate drawing will not be accepted. $ABC$ is a triangle with vertices $A(0, h)$, $B(2h, -2)$ and $C(7, 5)$, where $h$ is a positive integer. The area of the triangle is 20 units$^2$.

1. Apply the shoelace formula to find an expression for the area of triangle $ABC$.

Area $= \frac{1}{2} \left| \begin{matrix} 0 & 2h & 7 & 0 \\ h & -2 & 5 & h \end{matrix} \right|$
$= \frac{1}{2} |(0)(-2) + (2h)(5) + (7)(h) - (h)(2h) - (-2)(7) - (5)(0)|$
$= \frac{1}{2} |10h + 7h - 2h^2 + 14|$
$= \frac{1}{2} |-2h^2 + 17h + 14|$

2. Equate the area expression to 20 and set up the quadratic equations.

$\frac{1}{2} |-2h^2 + 17h + 14| = 20$
$|-2h^2 + 17h + 14| = 40$
This gives two cases:
Case 1: $-2h^2 + 17h + 14 = 40$
Case 2: $-2h^2 + 17h + 14 = -40$

3. Solve the quadratic equations and determine the valid integer value for $h$.

Case 1:
$2h^2 - 17h + 26 = 0$
$(2h - 13)(h - 2) = 0$
$h = 6.5$ or $h = 2$

Case 2:
$2h^2 - 17h - 54 = 0$
$h = \frac{17 \pm \sqrt{(-17)^2 - 4(2)(-54)}}{2(2)}$
$h = \frac{17 \pm \sqrt{721}}{4}$ (not an integer)

Since $h$ is given to be a positive integer, we reject $h = 6.5$ and the non-integer roots from Case 2.
Therefore, $h = 2$ (shown).

1. Find the coordinates of $A$ and $B$ using the value of $h$ from part (a)

Since $h = 2$, the coordinates are $A(0, 2)$ and $B(4, -2)$.

2. Find the midpoint of $AB$

Midpoint of $AB = \left(\frac{0 + 4}{2}, \frac{2 + (-2)}{2}\right) = (2, 0)$

3. Find the gradient of $AB$ and the gradient of the perpendicular bisector

Gradient of $AB = \frac{-2 - 2}{4 - 0} = \frac{-4}{4} = -1$

Let the gradient of the perpendicular bisector be $m_{\perp}$.
$m_{\perp} \times (-1) = -1 \implies m_{\perp} = 1$

4. Find the equation of the perpendicular bisector

Using the point $(2, 0)$ and gradient $1$:
$y - 0 = 1(x - 2)$
$y = x - 2$

1. Find the coordinates of $A$ and $B$ using $h=2$

Given $h = 2$, the vertices are $A(0, 2)$ and $B(4, -2)$. The vertex $C$ is $(7, 5)$.

2. Calculate the length of side $AC$

$AC = \sqrt{(7 - 0)^2 + (5 - 2)^2}$
$AC = \sqrt{7^2 + 3^2} = \sqrt{49 + 9} = \sqrt{58}$

3. Calculate the length of side $BC$ and conclude

$BC = \sqrt{(7 - 4)^2 + (5 - (-2))^2}$
$BC = \sqrt{3^2 + 7^2} = \sqrt{9 + 49} = \sqrt{58}$
Since $AC = BC$, triangle $ABC$ has two equal sides and is therefore isosceles.