In the coordinate plane, the line passing through points $(2023,0)$ and $(-2021,2024)$ also passes through $(1,c)$ for a constant $c$. Find $c$.

[i]Proposed by Andy Xu[/i]
1. To find the value of \( c \) for the point \((1, c)\) on the line passing through \((2023, 0)\) and \((-2021, 2024)\), we first need to determine the equation of the line.

2. The slope \( m \) of the line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
   \[
   m = \frac{y_2 - y_1}{x_2 - x_1}
   \]
   Substituting the given points \((2023, 0)\) and \((-2021, 2024)\):
   \[
   m = \frac{2024 - 0}{-2021 - 2023} = \frac{2024}{-4044} = -\frac{2024}{4044} = -\frac{1012}{2022} = -\frac{506}{1011} = -\frac{253}{505}
   \]

3. Using the point-slope form of the line equation \( y - y_1 = m(x - x_1) \), we can write the equation of the line. Using the point \((2023, 0)\):
   \[
   y - 0 = -\frac{253}{505}(x - 2023)
   \]
   Simplifying, we get:
   \[
   y = -\frac{253}{505}x + \frac{253}{505} \cdot 2023
   \]

4. To find the y-intercept when \( x = 1 \), substitute \( x = 1 \) into the line equation:
   \[
   c = -\frac{253}{505}(1) + \frac{253}{505} \cdot 2023
   \]
   Simplifying further:
   \[
   c = -\frac{253}{505} + \frac{253 \cdot 2023}{505}
   \]
   \[
   c = \frac{253 \cdot 2023 - 253}{505}
   \]
   \[
   c = \frac{253(2023 - 1)}{505}
   \]
   \[
   c = \frac{253 \cdot 2022}{505}
   \]
   \[
   c = \frac{253 \cdot 2 \cdot 1011}{505}
   \]
   \[
   c = \frac{506 \cdot 1011}{505}
   \]
   \[
   c = 1011 \cdot \frac{506}{505}
   \]
   \[
   c = 1011 \cdot 1 = 1012
   \]

5. Therefore, the value of \( c \) is:
   \[
   c = \boxed{1012}
   \]