A triangle has vertices $A=(3,0), B=(0,3)$, and $C$, where $C$ lies on the line with equation $x+y=7$. What is the area of this triangle?
Observe that the height $h$ relative to side $AB$ of all triangles $\triangle ABC$ that have vertex $C$ on the line $x+y=7$ is always the same, since the line $x+y=7$ is parallel to the line $x+y=3$ that passes through $A$ and $B$. Therefore, all these triangles have the same area, namely,

$$
\frac{1}{2}(AB \times h)
$$

It remains, therefore, to determine $AB$ and $h$. Since $AB$ is the hypotenuse of a right triangle that has both legs equal to 3, it follows from the Pythagorean Theorem that

$$
AB=\sqrt{3^{2}+3^{2}}=\sqrt{18}=3 \sqrt{2}
$$

On the other hand, $h$ is the distance between the parallel lines $x+y=3$ determined by points $A$ and $B$ and $x+y=7$. The line $x=y$ is perpendicular to these parallel lines and forms a $45^{\circ}$ angle with the $Ox$ axis. Let $M$ be the foot of the perpendicular to the line $x+y=7$ drawn from $A$ and $D=(7,0)$ the point of intersection of the line $x+y=7$ with the $Ox$ axis. The triangle $\triangle AMD$ thus formed is a right isosceles triangle, with two legs equal to $h$ and hypotenuse $7-3=4$. By the Pythagorean Theorem, it follows that $4^{2}=h^{2}+h^{2}=2 h^{2}$, that is, $h=\sqrt{8}=2 \sqrt{2}$. Thus, the area sought is given by

$$
\frac{1}{2}(3 \sqrt{2} \times 2 \sqrt{2})=6
$$