Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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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:
\[ (x, y) = \left( \begin{matrix}
\ \\
\ \\
\end{matrix} \](https://content.bartleby.com/qna-images/question/c3d3bee5-0c39-4b7b-a553-642953afa184/d0ec6207-8ba6-4be5-8ef1-589b916e12c5/0o5ro2m.png)
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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