2. (a) When $x=0$, the equation $2 x+3 y=12$ becomes $3 y=12$ and so $y=4$. Thus, the coordinates of $A$ are $(0,4)$.
When $y=0$, the equation $2 x+3 y=12$ becomes $2 x=12$ and so $x=6$. Thus, the coordinates of $B$ are $(6,0)$.
Thus, the vertices of $\triangle A O B$ are $A(0,4), O(0,0)$ and $B(6,0)$.
Since $\triangle A O B$ is right-angled at $O$, its area equals $\frac{1}{2} \times A O \times B O=\frac{1}{2} \times 4 \times 6=12$.
(b) When $x=0$, the equation $6 x+5 y=c$ becomes $5 y=c$ and so $y=\frac{1}{5} c$. Thus, the coordinates of $D$ are $\left(0, \frac{1}{5} c\right)$.
When $y=0$, the equation $6 x+5 y=c$ becomes $6 x=c$ and so $x=\frac{1}{6} c$. Thus, the coordinates of $E$ are $\left(\frac{1}{6} c, 0\right)$.
Thus, the vertices of $\triangle D O E$ are $D\left(0, \frac{1}{6} c\right), O(0,0)$ and $E\left(\frac{1}{2} c, 0\right)$.

Since $\triangle D O E$ is right-angled at $O$, its area equals $\frac{1}{2} \times D O \times E O=\frac{1}{2} \times \frac{1}{5} c \times \frac{1}{6} c$.
Since the area of $\triangle D O E$ is equal to 240 , we obtain $\frac{1}{2} \times \frac{1}{5} c \times \frac{1}{6} c=240$, which gives $\frac{1}{60} c^{2}=240$
Thus, $c^{2}=60 \times 240=14400$. Since $c>0$, then $c=120$.
(c) When $x=0$, the equation $(2 m) x+y=4 m$ becomes $y=4 m$. Thus, the coordinates of $P$ are $(0,4 m)$.
When $y=0$, the equation $(2 m) x+y=4 m$ becomes $(2 m) x=4 m$ and so $x=2$. (We note that $100 \leq mm$ and $7>4$. then $7 n>4 m$ and so $S$ is farther un the $u$-axis than $P$.

To calculate the area of quadrilateral $P Q R S$, we subtract the area of $\triangle P O Q$ from the area of $\triangle S O R$.
Therefore,
$$
\begin{array}{l}
2022=\frac{1}{2} \times S O \times R O-\frac{1}{2} \times P O \times Q O \\
2022=\frac{1}{2} \times 7 n \times 4-\frac{1}{2} \times 4 m \times 2 \\
2022=14 n-4 m \\
1011=7 n-2 m
\end{array}
$$

We want to find two pairs $(m, n)$ of positive integers with $100 \leq m<n$ and for which $7 n-2 m=1011$.
When $m=100$, we get $7 n-200=1011$ and so $7 n=1211$ which gives $n=173$. The pair $(m, n)=(100,173)$ satisfies the requirements.
From $7 \times 173-2 \times 100=1011$, we add and subtract 14 to obtain
$$
7 \times 173+14-2 \times 100-14=1011
$$
which gives
$$
7 \times 173+7 \times 2-2 \times 100-2 \times 7=1011
$$
and so
$$
7 \times 175-2 \times 107=1011
$$

Therefore, $(m, n)=(107,175)$ is a second pair that satisfies the given requirements.
2. A line that is neither horizontal nor vertical intersects the $y$-axis when $x=0$ and intersects the $x$-axis when $y=0$. For example, the line with equation $4 x+5 y=40$ intersects the $y$-axis at $(0,8)$ (because $4 \times 0+5 y=40$ gives $y=8$ ) and the $x$-axis at $(10,0)$ (because $4 x+5 \times 0=40$ gives $x=10$ ).
(a) The line with equation $2 x+3 y=12$ intersects the $y$-axis at $A$ and the $x$-axis at $B$. If $O$ is the origin, determine the area of $\triangle A O B$.
(b) Suppose that $c>0$. The line with equation $6 x+5 y=c$ intersects the $y$-axis at $D$ and the $x$-axis at $E$. If $O$ is the origin and the area of $\triangle D O E$ is 240 , determine the value of $c$.
(c) Suppose that $m$ and $n$ are integers with $100 \leq m$ and $m<n$. The line with equation $(2 m) x+y=4 m$ intersects the $y$-axis at $P$ and the $x$-axis at $Q$. The line with equation $(7 n) x+4 y=28 n$ intersects the $y$-axis at $S$ and the $x$-axis at $R$. If quadrilateral $P Q R S$ has area 2022, determine two possible pairs $(m, n)$.