The number of lattice points bound by the x and y axes, and the line $3x-y=24$.

Question

The right triangle bound by the x and y axes and the line $3x-y=6$ contains 2 lattice points in its interior. How many lattice points will be contained in the interior of a triangle bound by the x and y axes and the line $3x-y=24$?

So I first convert the line in question to slope intercept form, which is $y=3x-24$. Thus, this means that the line in question has a y-intercept of $-24$ and an x-intercept of $8$. This means that the line in question makes a right triangle with a dilation of $4$ from the origin in relation to the triangle the problem gives. I've heard something like that when a side is dilated with a factor of $r$, then the area is dilated with a factor of $r^2$. So I multiply $2$ by $4^2=16$ which gives me $32$. This is wrong, so what exactly is wrong with my reasoning? How can I solve this problem?

Answer 1

If you plot the line $3x - y = 24$, you will see that the triangular region in question is located in the fourth quadrant, namely $x > 0$, $y < 0$. So the number of lattice points in the interior of the triangle is equal to the number of ordered integer pairs $(x,y)$ with $x > 0$ and $y < 0$ that satisfy $3x - y < 24$. The direction of inequality is easily found by noting that the set of points we wish to count lies on the same side of the line $3x - y = 24$ as the origin $(0,0)$, thus when substituting $(x,y) = (0,0)$, the inequality must hold.

Now, when $y = 0$, $x < 8$, so the range of permissible $x$ values is $x \in \{1, 2, \ldots, 7\}$. For such an $x$, the range of permissible $y$ values is $0 > y > 3x-24$. The integers in this range are $\{3x-24 + 1, 3x-24+2, \ldots, -1\}$. How many are there? Clearly, there are $23-3x$ of them.

So the total number of lattice points in the interior of the triangle is $$\sum_{x=1}^7 (23-3x) = 77.$$

Answer 2

let:$A(0,0)$ ,$B(8,0)$ and$C(0,-24)$

the number of lattice points in the interior of the triangle ABC is equal to the number of ordered integer pairs $(x,y)$ with $x>0$ ,$y<0$ and $3x-24<y$

we pose $f(x)=3x-24$

$f(i-1)=3\times (i-1)-24=f(i)-3$

$f(8)=0$,$f(7)=-3$

the number of lattice points such that $-3<y<0$is equal:$2\times 7$

if $1\leq i \leq 7$ the number of lattice points such that $f(i-1)<y\leq f(i)$ is equal:$3 \times (i-1)$

So the total number of lattice points in the interior of the triangle is:

$2\times 7+\sum^2_{7}3\times (i-1)=2\times 7+\sum^6_{1}3\times i=77$

Answer 3

[Just a version of heropup's answer in more detail so I can understand]

If you plot the line $3x - y = 24$, you will see that the triangular region in question is located in the fourth quadrant, namely $x > 0$, $y < 0$. So the number of lattice points in the interior of the triangle is equal to the number of ordered integer pairs $(x,y)$ with $x > 0$ and $y < 0$ that satisfy $3x - y < 24$. The direction of inequality is easily found by noting that the set of points we wish to count lies on the same side of the line $3x - y = 24$ as the origin $(0,0)$, thus when substituting $(x,y) = (0,0)$, the inequality must hold.

Now, when $y = 0$, $x < 8$, so the range of permissible $x$ values is $x \in \{1, 2, \ldots, 7\}$. You know that $y$ is negative, and that it must satisfy $3x−y<24$, thus the range of permissible $y$ values is $0 > y > 3x-24$. The integers in this range are $\{3x-24 + 1, 3x-24+2, \ldots, -2, -1\}$. How many are there? If I said for example "how many integers are in the set $\{−1,−2,\ldots,−25\}$, you'd say $(-1)\cdot(-25)=25$. Using the same logic, there are $(-1)\cdot(3x-24+1)=23-3x$ of them. Then, we add this up for each possible value of $x$ in $\{1,2,\ldots,7\}$.

So the total number of lattice points in the interior of the triangle is $$\sum_{x=1}^7 (23-3x) = 77.$$