Problem 48
Question
Find the parametric equation of the line in the \(x-y\) plane that goes through the given points. Then eliminate the parameter to find the equation of the line in standard form. $$ (2,1) \text { and }(3,5) $$
Step-by-Step Solution
Verified Answer
The parametric equations are \(x = 2 + t\), \(y = 1 + 4t\). The line in standard form is \(4x - y = 7\).
1Step 1: Find the Direction Vector
To find the parametric equation of a line, we first need a direction vector. Given two points, \((x_1, y_1)\) and \((x_2, y_2)\), the direction vector \(\mathbf{d}\) can be found as \((x_2 - x_1, y_2 - y_1)\). Substitute the points \((2, 1)\) and \((3, 5)\) into the equation: \(\mathbf{d} = (3 - 2, 5 - 1) = (1, 4)\).
2Step 2: Write the Parametric Equations
With a starting point \((2,1)\) and a direction vector \((1,4)\), the parametric equations for the line can be formulated as \(x = 2 + t\) and \(y = 1 + 4t\), where \(t\) is the parameter.
3Step 3: Eliminate the Parameter
To find the equation of the line in standard form, eliminate the parameter \(t\) from the parametric equations. Solve for \(t\) in terms of \(x\): \(x = 2 + t \Rightarrow t = x - 2\). Substitute \(t\) in the equation for \(y\): \(y = 1 + 4(x - 2)\).
4Step 4: Simplify the Expression
Distribute the 4 and then simplify: \(y = 1 + 4x - 8\), which simplifies to \(y = 4x - 7\).
5Step 5: Write the Equation in Standard Form
The standard form of a line is \(Ax + By = C\). Rearrange the simplified equation to \(4x - y = 7\), which is the standard form of the line.
Key Concepts
Direction VectorStandard FormEliminate Parameter
Direction Vector
To understand how to find the parametric equation of a line, it's important to start with the direction vector. A direction vector aims to point out the direction in which the line extends. This vector is derived using two given points that lie on the line, in this case, (2,1) and (3,5).
The direction vector \(\mathbf{d} = (x_2 - x_1, y_2 - y_1)\) is calculated by subtracting the corresponding coordinates of these two points. For the given points, the calculation becomes:
The direction vector \(\mathbf{d} = (x_2 - x_1, y_2 - y_1)\) is calculated by subtracting the corresponding coordinates of these two points. For the given points, the calculation becomes:
- The direction vector is \(\mathbf{d} = (3 - 2, 5 - 1) = (1, 4)\)
Standard Form
When we need the standard form of a line's equation, we aim for a neat, well-recognized format. The standard form appears as \(Ax + By = C\). This form has specific coefficients and constants where both \(A\) and \(B\) should ideally be integers.
In this exercise, after framing the line with parametric equations using a start point and a direction vector, we move to convert this into standard form.
In this exercise, after framing the line with parametric equations using a start point and a direction vector, we move to convert this into standard form.
- Start off with the equation in parametric terms, such as \(x = 2 + t\) and \(y = 1 + 4t\).
- Rearrange the parametric equation to eliminate the parameter \(t\) (which we will discuss in the next section) and simplify it to get \(y = 4x - 7\).
- To change this into the standard form, rearrange it to \(4x - y = 7\).
Eliminate Parameter
Eliminating the parameter \(t\) is a key step in transitioning from a parametric equation to a coordinate-based, standard form. The process involves expressing \(t\) in terms of one coordinate, and then substituting it into the other equation. Let's break it down for clarity.
- Start with the parametric equations: \(x = 2 + t\) and \(y = 1 + 4t\).
- Isolate \(t\) in one equation. From \(x = 2 + t\), we get \(t = x - 2\).
- Substitute \(t\) from this expression into the equation for \(y\). This becomes \(y = 1 + 4(x - 2)\).
- Finally, simplify the expression: expand and combine terms to get \(y = 4x - 7\).
Other exercises in this chapter
Problem 47
Use a rotation matrix to rotate the vector \(\left[\begin{array}{r}5 \\\ -3\end{array}\right]\) clockwise by the angle \(\pi / 7\).
View solution Problem 48
$$ A=\left[\begin{array}{rr} -1 & 1 \\ 2 & 3 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & -2 \\ 3 & 2 \end{array}\right] $$ Show that \((A B)^{-1}=B^
View solution Problem 49
$$ A=\left[\begin{array}{rr} -1 & 1 \\ 2 & 3 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & -2 \\ 3 & 2 \end{array}\right] $$ Find the inverse (if it e
View solution Problem 49
Find the parametric equation of the line in the \(x-y\) plane that goes through the given points. Then eliminate the parameter to find the equation of the line
View solution