Problem 50 - Calculus for Biology and Medicine

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. \((2,3)\) and \((-1,-4)\)

Step-by-Step Solution

Answer

Parametric equation: \(x = 2 - 3t, y = 3 - 7t\); Standard form: \(7x - 3y = 5\).

Step 1: Find the Direction Vector

To find the parametric equation of the line going through the points \((2, 3)\) and \((-1, -4)\), we first calculate the direction vector. Subtract the coordinates of the two points: \((-1 - 2, -4 - 3) = (-3, -7)\). This gives us the direction vector \((-3, -7)\).

Step 2: Write the Parametric Equations

Use one of the points, say \((2, 3)\), and the direction vector \((-3, -7)\) to write the parametric equations of the line. Let \(t\) be the parameter. The parametric equations are: \(x = 2 - 3t\) and \(y = 3 - 7t\).

Step 3: Eliminate the Parameter

To eliminate the parameter \(t\), solve one of the parametric equations for \(t\). Let's solve \(x = 2 - 3t\) for \(t\):\[ 3t = 2 - x \]\[ t = \frac{2-x}{3} \]Substitute \(t\) in the second equation:\[ y = 3 - 7\left(\frac{2-x}{3}\right) \]Simplify:\[ y = 3 - \frac{14 - 7x}{3} \]\[ y = 3 - \frac{14}{3} + \frac{7x}{3} \]\[ y = \frac{9}{3} - \frac{14}{3} + \frac{7x}{3} \]\[ y = \frac{7x}{3} - \frac{5}{3} \]

Step 4: Convert to Standard Form

Multiply through by 3 to eliminate the fractions:\[ 3y = 7x - 5 \]Rearrange to get the standard form \(Ax + By = C\):\[ 7x - 3y = 5 \].

Key Concepts

Direction Vector

The direction vector is a crucial concept when dealing with parametric equations. It provides the line's orientation in space, which means it shows us the direction in which the line extends. To find this vector, we subtract the coordinates of two given points on the line.
In our exercise, we have the points \(2, 3\) and \(-1, -4\). Subtracting these, we calculate the direction vector as \(-1 - 2, -4 - 3\) which gives \(-3, -7\).

• Step 1: Take the difference of the x-coordinates: -1 - 2 = -3
• Step 2: Take the difference of the y-coordinates: -4 - 3 = -7
• The direction vector is then \((-3, -7)\)

This direction vector helps to form the parametric equations by combining with any point on the line.

Eliminating Parameter

Eliminating the parameter from a set of parametric equations is a key step when converting these equations into a standard form equation. This process involves expressing the parameter in terms of one variable and substituting it into the other equation.
For example, given the parametric equations \(x = 2 - 3t\) and \(y = 3 - 7t\), we start by solving \(x = 2 - 3t\) for \(t\):

• Rearrange to: \(3t = 2 - x\)
• Solve for \(t\): \(t = \frac{2-x}{3}\)

Next, substitute \(t\) in the equation for \(y\):
Substitute \(t = \frac{2-x}{3}\) into \(y = 3 - 7t\):

• \(y = 3 - 7\left(\frac{2-x}{3}\right)\) simplifies to \(y = \frac{7x}{3} - \frac{5}{3}\)

This eliminates the parameter \(t\), yielding an equation that relates \(y\) and \(x\) directly.

Standard Form Equation

The standard form of a line's equation has the structure \(Ax + By = C\), where \(A\), \(B\), and \(C\) are integers, and \(A\) should ideally be non-negative. Transitioning from parametric equations to this form involves standard algebraic manipulation.
From our rearranged equation \(y = \frac{7x}{3} - \frac{5}{3}\), we begin by multiplying every term by 3 to eliminate fractions:

• Multiply: \(3y = 7x - 5\)
• Rearrange: \(7x - 3y = 5\)

This is now in the desired standard form, \(7x - 3y = 5\).
In this linear equation:

• \(A = 7\)
• \(B = -3\)
• \(C = 5\)

The standard form is useful for easy identification of intercepts and direct applications in further algebraic contexts.