5. There are five pairs of integers $(a, b)$ with $0 \leq a \leq 10$ and $0 \leq b \leq 10$ for which the points $P(1,1), Q(4,5)$ and $R(a, b)$ form a triangle that has area 6 . What are these five pairs of integers $(a, b)$ ?
5. Consider the point $R(4,1)$.
Here, $P R$ is horizontal and $R Q$ is vertical and so $\triangle P Q R$ is right-angled at $R$.
Because $P R=4-1=3$ and $R Q=5-1=4$, then the area of $\triangle P Q R$ is $\frac{1}{2} \times P R \times R Q$ which equals $\frac{1}{2} \times 3 \times 4$ or 6 .
Therefore, the point $R(4,1)$ has the property that $\triangle P Q R$ has area 6 .
If a point $X$ lies on the line through $R$ parallel to $P Q$, then the perpendicular distance from $X$ to the line through $P Q$ is the same as the perpendicular distance from $R$ to $P Q$. This means that the height of $\triangle P Q X$ equals the height of $\triangle P Q R$ and so the areas of these triangles are equal.

In other words, any point $X$ on the line through $R$ parallel to $P Q$ has the property that the area of $\triangle P Q X$ is 6 .
To get from $P$ to $Q$, we move 3 units to the right and 4 units up. Therefore, starting at $R(4,1)$ and moving to the points $(4+3,1+4)=(7,5)$ will get to another point on the line through $R(4,1)$ parallel to $P Q$ (since moving right 3 and up 4 preserves the
slope of $\frac{4}{3}$.).
Similarly, the point $(7+3,5+4)=(10,9)$ is on this line.
In other words, the triangle with vertices at $(1,1),(4,5)$ and $(7,5)$ and the triangle with vertices at $(1,1),(4,5)$ and $(10,9)$ both have areas of 6 .
(To do this more formally, we could determine that the equation of the line through $R$ parallel to $P Q$ is $y=\frac{4}{3} x-\frac{13}{3}$ and then check the possible integer $x$-coordinates of points on this line (between 0 and 10, inclusive) to get the points $(1,-3),(4,1),(7,5),(10,9)$, omitting $(1,-3)$ since -3 is not in the correct range).
Next, consider $S(1,5)$.
Again, the area of $\triangle P Q S$ is 6 because $P S$ is vertical (with length 4) and $Q S$ is horizontal (with length 3 ).
Moving right 3 and up 4 from $(1,5)$ gives the point $(4,9)$.
Since we are told that there are five such points, then we have found all of the points and they are $(4,1),(7,5),(10,9),(1,5)$, and $(4,9)$.
(We could note that any point $T$ for which the area of $\triangle P Q T$ is 6 must lie on one of these two lines. This is because $T$ must have a particular perpendicular distance to the line through $P Q$ and must be on one side of this line or the other. Because these two lines are fixed, there are indeed only five points with these properties.)
ANSWER: $(4,1),(7,5),(10,9),(1,5)$, and $(4,9)$