Solutions to this question by accurate drawing will NOT be accepted. The figure shows a right-angled triangle $ABC$, where points $A$, $B$ and $C$ are $(k, 0)$, $(2, 8)$ and $(-2, -4)$ respectively. $BC$ cuts the $y$-axis at $N$. $M$ is a point on $AC$.

1. Express the gradients of $AB$ and $AC$ in terms of $k$

Gradient of $AB = \frac{8 - 0}{2 - k} = \frac{8}{2 - k}$
Gradient of $AC = \frac{-4 - 0}{-2 - k} = \frac{-4}{-2 - k} = \frac{4}{k + 2}$

2. Form an equation using the condition for perpendicular lines

Since $AB \perp AC$, the product of their gradients is $-1$:
$\left(\frac{8}{2 - k}\right) \left(\frac{4}{k + 2}\right) = -1$
$\frac{32}{(2 - k)(2 + k)} = -1$

3. Solve the equation for $k$ and reject the invalid value

$32 = -(4 - k^2)$
$32 = -4 + k^2$
$k^2 = 36$
$k = 6 \text{ or } k = -6$
From the figure, point $A$ lies on the positive $x$-axis, so $k > 0$.
Therefore, $k = 6$.

1. Find the equation of the line $BC$ to determine the coordinates of $N$.

Gradient of $BC = \frac{8 - (-4)}{2 - (-2)} = \frac{12}{4} = 3$
Equation of $BC$: $y - 8 = 3(x - 2)$
$y = 3x + 2$
Since $N$ lies on the $y$-axis, $x = 0$, so $y = 2$.
The coordinates of $N$ are $(0, 2)$.

2. Calculate the midpoint of $BC$ and show it is $N$.

Midpoint of $BC = \left( \frac{2 + (-2)}{2}, \frac{8 + (-4)}{2} \right)$
$= \left( \frac{0}{2}, \frac{4}{2} \right)$
$= (0, 2)$
Since the coordinates of $N$ are identical to the midpoint of $BC$, $N$ is the midpoint of $BC$.

1. Recall or calculate the coordinates of $A$ and $N$ from earlier parts

From part (i), $A = (6, 0)$.
From part (ii), $N = (0, 2)$.
We are given $B = (2, 8)$ and $C = (-2, -4)$.

2. Calculate the area of triangle $ABC$

Using the shoelace formula for $\triangle ABC$ with vertices $(6, 0)$, $(2, 8)$, and $(-2, -4)$:
$\text{Area} = \frac{1}{2} |(6)(8) + (2)(-4) + (-2)(0) - (0)(2) - (8)(-2) - (-4)(6)|$
$\text{Area} = \frac{1}{2} |48 - 8 + 0 - 0 + 16 + 24| = \frac{1}{2} |80| = 40 \text{ units}^2$

3. Find the area of triangle $MNC$

The area of quadrilateral $ABNM$ is given as $25 \text{ units}^2$.
$\text{Area of } \triangle MNC = \text{Area of } \triangle ABC - \text{Area of } ABNM$
$\text{Area of } \triangle MNC = 40 - 25 = 15 \text{ units}^2$

4. Find the equation of the line $AC$ to express the coordinates of $M$

Gradient of $AC = \frac{-4 - 0}{-2 - 6} = \frac{-4}{-8} = \frac{1}{2}$.
Equation of $AC$ using point $A(6, 0)$:
$y - 0 = \frac{1}{2}(x - 6) \implies y = \frac{1}{2}x - 3$
Let the coordinates of $M$ be $(x, \frac{1}{2}x - 3)$.

5. Set up an equation for the area of triangle $MNC$ and solve for $x$

Using the shoelace formula for $\triangle MNC$ with vertices $M(x, \frac{1}{2}x - 3)$, $N(0, 2)$, and $C(-2, -4)$:
$\text{Area} = \frac{1}{2} |x(2) + 0(-4) + (-2)(\frac{1}{2}x - 3) - (\frac{1}{2}x - 3)(0) - (2)(-2) - (-4)(x)| = 15$
$\frac{1}{2} |2x + 0 - x + 6 - 0 + 4 + 4x| = 15$
$\frac{1}{2} |5x + 10| = 15$
$|5x + 10| = 30$
$5x + 10 = 30 \quad \text{or} \quad 5x + 10 = -30$
$5x = 20 \implies x = 4$
$5x = -40 \implies x = -8$

6. Determine the correct coordinates of $M$

Since $M$ lies on the line segment $AC$, its $x$-coordinate must be between the $x$-coordinates of $C(-2)$ and $A(6)$.
Therefore, $x = 4$.
Substitute $x = 4$ into the equation of $AC$:
$y = \frac{1}{2}(4) - 3 = -1$
The coordinates of $M$ are $(4, -1)$.