Find the point of intersection of the lines r;(t) = (-1, 1) + t(8, 4) and rɔ(s) = (2, 1) + s(8, 6) in R?. (х, у) -

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Title: Finding the Point of Intersection of Two Lines in R²**

**Problem Statement:**
The goal is to find the point of intersection of the lines defined by the parametric equations:
\[ \mathbf{r_1}(t) = \langle -1, 1 \rangle + t \langle 8, 4 \rangle \]
and
\[ \mathbf{r_2}(s) = \langle 2, 1 \rangle + s \langle 8, 6 \rangle \]
in \(\mathbb{R}^2\).

**Expression for Point of Intersection:**
\[ (x, y) = \left( \text{input box} \right) \]

**Explanation:**
We need to determine the values of \(t\) and \(s\) for which the two parametric equations yield the same coordinates \((x, y)\).

1. **Equation (1) from \(\mathbf{r_1}(t)\):**
   \[
   x_1(t) = -1 + 8t
   \]
   \[
   y_1(t) = 1 + 4t
   \]

2. **Equation (2) from \(\mathbf{r_2}(s)\):**
   \[
   x_2(s) = 2 + 8s
   \]
   \[
   y_2(s) = 1 + 6s
   \]

To find the intersection point, set \(x_1(t) = x_2(s)\) and \(y_1(t) = y_2(s)\):
\[
-1 + 8t = 2 + 8s  \tag{1}
\]
\[
1 + 4t = 1 + 6s  \tag{2}
\]

Solving these equations will yield the values of \(t\) and \(s\) for which the lines intersect. The corresponding coordinates can then be obtained by substituting \(t\) back into \(\mathbf{r_1}(t)\) or \(\mathbf{r_2}(s)\).

**Submission Box:**
Here, the coordinates of the intersection point can be entered:
\[ (x, y) = \left( \begin{matrix}
\ \\
\ \\
\end{matrix} \
Transcribed Image Text:**Title: Finding the Point of Intersection of Two Lines in R²** **Problem Statement:** The goal is to find the point of intersection of the lines defined by the parametric equations: \[ \mathbf{r_1}(t) = \langle -1, 1 \rangle + t \langle 8, 4 \rangle \] and \[ \mathbf{r_2}(s) = \langle 2, 1 \rangle + s \langle 8, 6 \rangle \] in \(\mathbb{R}^2\). **Expression for Point of Intersection:** \[ (x, y) = \left( \text{input box} \right) \] **Explanation:** We need to determine the values of \(t\) and \(s\) for which the two parametric equations yield the same coordinates \((x, y)\). 1. **Equation (1) from \(\mathbf{r_1}(t)\):** \[ x_1(t) = -1 + 8t \] \[ y_1(t) = 1 + 4t \] 2. **Equation (2) from \(\mathbf{r_2}(s)\):** \[ x_2(s) = 2 + 8s \] \[ y_2(s) = 1 + 6s \] To find the intersection point, set \(x_1(t) = x_2(s)\) and \(y_1(t) = y_2(s)\): \[ -1 + 8t = 2 + 8s \tag{1} \] \[ 1 + 4t = 1 + 6s \tag{2} \] Solving these equations will yield the values of \(t\) and \(s\) for which the lines intersect. The corresponding coordinates can then be obtained by substituting \(t\) back into \(\mathbf{r_1}(t)\) or \(\mathbf{r_2}(s)\). **Submission Box:** Here, the coordinates of the intersection point can be entered: \[ (x, y) = \left( \begin{matrix} \ \\ \ \\ \end{matrix} \
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