For every rational number $m>0$ we consider the function $f_m:\mathbb{R}\rightarrow\mathbb{R},f_m(x)=\frac{1}{m}x+m$. Denote by $G_m$ the graph of the function $f_m$. Let $p,q,r$ be positive rational numbers.

a) Show that if $p$ and $q$ are distinct then $G_p\cap G_q$ is non-empty.

b) Show that if $G_p\cap G_q$ is a point with integer coordinates, then $p$ and $q$ are integer numbers.

c) Show that if $p,q,r$ are consecutive natural numbers, then the area of the triangle determined by intersections of $G_p,G_q$ and $G_r$ is equal to $1$.
### Part (a)
1. Consider the functions \( f_p(x) = \frac{1}{p}x + p \) and \( f_q(x) = \frac{1}{q}x + q \).
2. To find the intersection \( G_p \cap G_q \), we need to solve \( f_p(x) = f_q(x) \):
   \[
   \frac{1}{p}x + p = \frac{1}{q}x + q
   \]
3. Rearrange the equation to isolate \( x \):
   \[
   \frac{1}{p}x - \frac{1}{q}x = q - p
   \]
   \[
   x \left( \frac{1}{p} - \frac{1}{q} \right) = q - p
   \]
4. Since \( p \neq q \), \( \frac{1}{p} - \frac{1}{q} \neq 0 \). Thus, we can solve for \( x \):
   \[
   x = \frac{q - p}{\frac{1}{p} - \frac{1}{q}} = \frac{q - p}{\frac{q - p}{pq}} = pq
   \]
5. Substitute \( x = pq \) back into either \( f_p(x) \) or \( f_q(x) \) to find \( y \):
   \[
   y = f_p(pq) = \frac{1}{p}(pq) + p = q + p
   \]
6. Therefore, the intersection point is \( (pq, p + q) \), which shows that \( G_p \cap G_q \) is non-empty.

### Part (b)
1. Given that \( G_p \cap G_q \) is a point with integer coordinates, let the intersection point be \( (a, b) \) where \( a \) and \( b \) are integers.
2. From part (a), we know the intersection point is \( (pq, p + q) \).
3. Since \( pq \) and \( p + q \) are integers, and \( p \) and \( q \) are rational numbers, we can write \( p = \frac{m}{n} \) and \( q = \frac{k}{l} \) where \( m, n, k, l \) are integers.
4. Then:
   \[
   pq = \frac{m}{n} \cdot \frac{k}{l} = \frac{mk}{nl}
   \]
   \[
   p + q = \frac{m}{n} + \frac{k}{l} = \frac{ml + kn}{nl}
   \]
5. For \( pq \) to be an integer, \( nl \) must divide \( mk \). For \( p + q \) to be an integer, \( nl \) must divide \( ml + kn \).
6. Since \( p \) and \( q \) are rational and the coordinates are integers, the simplest case is when \( n = l = 1 \), making \( p \) and \( q \) integers.

### Part (c)
1. Let \( p = n \), \( q = n + 1 \), and \( r = n + 2 \) where \( n \) is a natural number.
2. The vertices of the triangle formed by the intersections of \( G_p \), \( G_q \), and \( G_r \) are:
   \[
   A = (pq, p + q) = (n(n+1), n + (n+1)) = (n^2 + n, 2n + 1)
   \]
   \[
   B = (pr, p + r) = (n(n+2), n + (n+2)) = (n^2 + 2n, 2n + 2)
   \]
   \[
   C = (qr, q + r) = ((n+1)(n+2), (n+1) + (n+2)) = (n^2 + 3n + 2, 2n + 3)
   \]
3. The area of the triangle with vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) is given by:
   \[
   \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
   \]
4. Substitute the coordinates of \( A \), \( B \), and \( C \):
   \[
   \text{Area} = \frac{1}{2} \left| (n^2 + n)((2n + 2) - (2n + 3)) + (n^2 + 2n)((2n + 3) - (2n + 1)) + (n^2 + 3n + 2)((2n + 1) - (2n + 2)) \right|
   \]
   \[
   = \frac{1}{2} \left| (n^2 + n)(-1) + (n^2 + 2n)(2) + (n^2 + 3n + 2)(-1) \right|
   \]
   \[
   = \frac{1}{2} \left| -n^2 - n + 2n^2 + 4n - n^2 - 3n - 2 \right|
   \]
   \[
   = \frac{1}{2} \left| -2 \right| = \frac{1}{2} \cdot 2 = 1
   \]

The final answer is \( \boxed{ 1 } \)