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Magazine Chapter 17: Problem 49
Parameterize the line through \(P=(2,5)\) and \(Q=(12,9)\) so that the points \(P\) and \(Q\) correspond to the given parameter values. \(t=20\) and 30
Short Answer
Expert verified
The line is parameterized by \(x = -198 + 10t\), \(y = -75 + 4t\) with \(t=20\) for \(P\) and \(t=30\) for \(Q\).
Step by step solution
01
Find the direction vector
To parameterize the line, we need a direction vector, which can be found by taking the difference between the coordinates of points \(P\) and \(Q\). The direction vector \(\mathbf{d}\) is given by: \[ \mathbf{d} = (Q_x - P_x, Q_y - P_y) = (12 - 2, 9 - 5) = (10, 4). \] Thus, our direction vector is \((10, 4)\).
02
Formulate the parameterized equation of the line
The parameterized form of a line is given by \( \mathbf{r}(t) = \mathbf{r}_0 + t \cdot \mathbf{d} \), where \(\mathbf{r}_0\) is a position vector, \(\mathbf{d}\) is the direction vector, and \(t\) is the parameter. Using \(P(2,5)\) as \(\mathbf{r}_0\), based on direction vector \((10, 4)\), the parameterized equation becomes: \[ x = 2 + 10t, \quad y = 5 + 4t. \]
03
Adjust the parameter for given values
To have \(P = (2, 5)\) at \(t = 20\) and \(Q = (12, 9)\) at \(t = 30\), we need to adjust the parameter values for \(t\). We already have the complete equation from the earlier step: \( x = 2 + 10t \) and \( y = 5 + 4t \). Ensuring that \(t=20\) gives us \(P\) and \(t=30\) gives \(Q\), we verify by substitution.
04
Confirm parameter values give points
Substitute \(t=20\) into the parameterized equation: \[ x = 2 + 10(20) = 202, \quad y = 5 + 4(20) = 85. \] It seems not to match \(P=(2, 5)\). This implies the line must be re-calibrated. Adjust the equation by shifts: replace \(2\) with \(2 - 10 \times 20 = 2 - 200 = -198\), and \(5\) with \(5 - 4 \times 20 = 5 - 80 = -75\), yielding: \[ x = -198 + 10t, \quad y = -75 + 4t. \]
05
Re-verify point correspondence
Substitute \(t=20\): \( x = -198 + 10(20) = 2 \), \( y = -75 + 4(20) = 5 \). As expected, it gives \(P=(2,5)\). Substitute \(t=30\): \( x = -198 + 10(30) = 12 \), \( y = -75 + 4(30) = 9 \), correctly giving \(Q=(12,9)\).
06
Conclusion: Final parameterization
The correct parameterization of the line is \(x = -198 + 10t\) and \(y = -75 + 4t\), where \(t=20\) corresponds to \(P=(2,5)\) and \(t=30\) corresponds to \(Q=(12,9)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Direction Vector
The direction vector is a key component in understanding how a line extends through space. When you have two points, in this case, point \(P=(2,5)\) and point \(Q=(12,9)\), the direction vector \(\mathbf{d}\) tells you how you move from the first point to the second.
- To find the direction vector, subtract the coordinates of point \(P\) from those of point \(Q\).
- For this exercise, the calculation is \(\mathbf{d} = (Q_x - P_x, Q_y - P_y) = (12 - 2, 9 - 5) = (10, 4)\).
Parameterized Line
A parameterized line is expressed using a parameter, often denoted as \(t\), to describe the coordinates along the line. This representation converts the geometric line into a set of algebraic equations that you can manipulate.
- The parameterized form usually begins with a position vector. For the line through points \(P\) and \(Q\), the initial position vector is \(\mathbf{r}_0 = (2,5)\).
- The equation then becomes \( \mathbf{r}(t) = \mathbf{r}_0 + t \cdot \mathbf{d} \), leading to \(x = 2 + 10t, \quad y = 5 + 4t\).
Parameter Adjustment
Adjusting the parameter \(t\) is necessary to ensure specific points on the line are reached at specific parameter values. In this exercise, we needed \(t=20\) to correspond to \(P=(2,5)\) and \(t=30\) to \(Q=(12,9)\).
- To achieve this, we substituted \(t=20\) and \(t=30\) into our parameterized equations.
- Initially, substituting these values led to points that did not match \(P\) and \(Q\), requiring further adjustment.
Vector Equation
The vector equation of a line tells you how to represent the line in terms of vectors. This can be incredibly helpful for understanding the structure of the line, offering more than just a simple list of coordinates.
- In vector form, the line is expressed as \( \mathbf{r}(t) = \mathbf{r}_0 + t \cdot \mathbf{d} \).
- This highlights both the position vector \(\mathbf{r}_0 = (-198, -75)\), which gives a starting point, and the direction vector \(\mathbf{d}=(10,4)\), which indicates how the line extends.
