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Let's solve each part of the problem given the parametric equations of the line \( r \):

\[
\begin{cases}
x = 2 - t \\
y = 1 + 4t
\end{cases}
\]

where \( t \in \mathbb{R} \).

### a) ¿Los puntos \( P(1,5) \) y \( Q(3,-2) \) pertenecen a \( r \)?

To determine if the points \( P(1,5) \) and \( Q(3,-2) \) belong to the line \( r \), we need to find a value of \( t \) such that both coordinates of the points satisfy the parametric equations.

1. **For point \( P(1,5) \):**
   - Set \( x = 1 \): 
     \[
     1 = 2 - t \implies t = 1
     \]
   - Check if \( y = 5 \) when \( t = 1 \):
     \[
     y = 1 + 4 \cdot 1 = 5
     \]
   - Therefore, \( P(1,5) \) belongs to \( r \).

2. **For point \( Q(3,-2) \):**
   - Set \( x = 3 \): 
     \[
     3 = 2 - t \implies t = -1
     \]
   - Check if \( y = -2 \) when \( t = -1 \):
     \[
     y = 1 + 4 \cdot (-1) = -3
     \]
   - Therefore, \( Q(3,-2) \) does not belong to \( r \).

### b) ¿Para qué valor del parámetro \( t \) se obtiene el punto \((-2,17)\)?

We need to find \( t \) such that the coordinates match \((-2,17)\).

1. Set \( x = -2 \):
   \[
   -2 = 2 - t \implies t = 4
   \]
2. Check \( y = 17 \) when \( t = 4 \):
   \[
   y = 1 + 4 \cdot 4 = 17
   \]
Thus, the value of \( t \) is \( 4 \).

### c) ¿Para qué valores de \( t \) se obtienen los puntos del segmento determinado por las intersecciones de la recta con los ejes coordenados?

To find the intersections with the axes, set \( x = 0 \) and \( y = 0 \).

1. **Intersection with the \(x\)-axis (\( y = 0 \)):**
   - Set \( y = 0 \):
     \[
     0 = 1 + 4t \implies t = -\frac{1}{4}
     \]
   - Corresponding \( x \):
     \[
     x = 2 - \left(-\frac{1}{4}\right) = 2.25
     \]

2. **Intersection with the \(y\)-axis (\( x = 0 \)):**
   - Set \( x = 0 \):
     \[
     0 = 2 - t \implies t = 2
     \]
   - Corresponding \( y \):
     \[
     y = 1 + 4 \cdot 2 = 9
     \]

The segment between these points corresponds to \( t \in \left[-\frac{1}{4}, 2\right] \).

### d) Calcule el área del triángulo que forma la recta con los ejes coordenados.

The vertices of the triangle are:
- Intersection with the \(x\)-axis: \(\left(2.25, 0\right)\)
- Intersection with the \(y\)-axis: \(\left(0, 9\right)\)
- The origin: \((0, 0)\)

The area \( A \) of the triangle is:
\[
A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.25 \times 9 = 10.125
\]

### e) Escriba otras ecuaciones paramétricas de la misma recta.

We can reparametrize the equations using a new parameter \( s = t - 1 \). The new parametric equations can be:

\[
\begin{cases}
x = 3 - s \\
y = -3 + 4s
\end{cases}
\]

### Follow-up Questions:

1. Can we find the Cartesian equation of the line from the given parametric form?
2. What is the slope of the line represented by the parametric equations?
3. How would the parametric equations change if the line passed through a different point?
4. Can we express the line in vector form instead of parametric?
5. How does the value of \( t \) influence the position on the line?

**Tip:** To find if a point lies on a line represented parametrically, solve for \( t \) in one equation and substitute into the other to check consistency.

Sean las ecuaciones paramétricas de una recta r dadas por: x = 2 - t, y = 1 + 4t, donde t ∈ ℝ. Determine si los puntos P(1,5) y Q(3,-2) pertenecen a la recta r, para qué valor del parámetro t se obtiene el punto (-2,17), los valores de t para los puntos de intersección con los ejes, el área del triángulo formado con los ejes, y otras ecuaciones paramétricas de la misma recta.

Explore how to solve problems involving parametric equations of a line, including finding points of intersection, calculating the area of a triangle formed with coordinate axes, and reparameterizing the same line. Suitable for students in Grades 11-12.

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