The points $J(2,7), K(5,3)$ and $L(r, t)$ form a triangle whose area is less than or equal to 10 . Let $\mathcal{R}$ be the region formed by all such points $L$ with $0 \leq r \leq 10$ and $0 \leq t \leq 10$. When written as a fraction in lowest terms, the area of $\mathcal{R}$ is equal to $\frac{300+a}{40-b}$ for some positive integers $a$ and $b$. The value of $a+b$ is
(A) 82
(B) 71
(C) 60
(D) 49
(E) 93
The distance between $J(2,7)$ and $K(5,3)$ is equal to $\sqrt{(2-5)^{2}+(7-3)^{2}}=\sqrt{3^{2}+4^{2}}=5$.

Therefore, if we consider $\triangle J K L$ as having base $J K$ and height $h$, then we want $\frac{1}{2} \cdot J K \cdot h \leq 10$ which means that $h \leq 10 \cdot \frac{2}{5}=4$.

In other words, $L(r, t)$ can be any point with $0 \leq r \leq 10$ and $0 \leq t \leq 10$ whose perpendicular distance to the line through $J$ and $K$ is at most 4 .

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-035.jpg?height=455&width=458&top_left_y=957&top_left_x=888)

The slope of the line through $J(2,7)$ and $K(5,3)$ is equal to $\frac{7-3}{2-5}=-\frac{4}{3}$.

Therefore, this line has equation $y-7=-\frac{4}{3}(x-2)$.

Muliplying through by 3 , we obtain $3 y-21=-4 x+8$ or $4 x+3 y=29$.

We determine the equation of the line above this line that is parallel to it and a perpendicular distance of 4 from it.

The equation of this line will be of the form $4 x+3 y=c$ for some real number $c$, since it is parallel to the line with equation $4 x+3 y=29$.

To determine the value of $c$, we determine the coordinates of one point on this line.

To determine such a point, we draw a perpendicular of length 4 from $K$ to a point $P$ above the line.

Since $J K$ has slope $-\frac{4}{3}$ and $K P$ is perpendicular to $J K$, then $K P$ has slope $\frac{3}{4}$.

Draw a vertical line from $P$ and a horizontal line from $K$, meeting at $Q$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-035.jpg?height=447&width=461&top_left_y=2178&top_left_x=886)

Since $K P$ has slope $\frac{3}{4}$, then $P Q: Q K=3: 4$, which means that $\triangle K Q P$ is similar to a $3-4-5$ triangle.

Since $K P=4$, then $P Q=\frac{3}{5} K P=\frac{12}{5}$ and $Q K=\frac{4}{3} K P=\frac{16}{5}$.

Thus, the coordinates of $P$ are $\left(5+\frac{16}{5}, 3+\frac{12}{5}\right)$ or $\left(\frac{41}{5}, \frac{27}{5}\right)$.

Since $P$ lies on the line with equation $4 x+3 y=c$, then

$$
c=4 \cdot \frac{41}{5}+3 \cdot \frac{27}{5}=\frac{164}{5}+\frac{81}{5}=\frac{245}{5}=49
$$

and so the equation of the line parallel to $J K$ and 4 units above it is $4 x+3 y=49$.

In a similar way, we find that the line parallel to $J K$ and 4 units below it has equation $4 x+3 y=9$. (Note that $49-29=29-9$.

This gives us the following diagram:

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-036.jpg?height=461&width=458&top_left_y=951&top_left_x=888)

The points $L$ that satisfy the given conditions are exactly the points within the square, below the line $4 x+3 y=49$ and above the line $4 x+3 y=9$. In other words, the region $\mathcal{R}$ is the region inside the square and between these lines.

To find the area of $\mathcal{R}$, we take the area of the square bounded by the lines $x=0, x=10$, $y=0$, and $y=10$ (this area equals $10 \cdot 10$ or 100 ) and subtract the area of the two triangles inside the square and not between the lines.

The line with equation $4 x+3 y=9$ intersects the $y$-axis at ( 0,3$)$ (we see this by setting $x=0$ ) and the $x$-axis at $\left(\frac{9}{4}, 0\right)$ (we see this by setting $y=0$ ).

The line with equation $4 x+3 y=49$ intersects the line $x=10$ at $(10,3)$ (we see this by setting $x=10$ ) and the line $y=10$ at $\left(\frac{19}{4}, 10\right)$ (we see this by setting $y=10$ ).

The bottom triangle that is inside the square and outside $\mathcal{R}$ has area $\frac{1}{2} \cdot 3 \cdot \frac{9}{4}=\frac{27}{8}$.

The top triangle that is inside the square and outside $\mathcal{R}$ has horizontal base of length $10-\frac{19}{4}$ or $\frac{21}{4}$ and vertical height of length $10-3$ or 7 , and thus has area $\frac{1}{2} \cdot \frac{21}{4} \cdot 7=\frac{147}{8}$.

Finally, this means that the area of $\mathcal{R}$ is

$$
100-\frac{27}{8}-\frac{147}{8}=100-\frac{174}{8}=100-\frac{87}{4}=\frac{313}{4}
$$

which is in lowest terms since the only divisors of the denominator that are larger than 1 are 2 and 4 , while the numerator is odd.

When we write this area in the form $\frac{300+a}{40-b}$ where $a$ and $b$ are positive integers, we obtain $a=13$ and $b=36$, giving $a+b=49$.

ANSwER: (D)