Problem 8 - Calculus for Biology and Medicine

Question

Determine the equation of the line that satisfies the stated requirements. Put the equation in standard form. The line passing through \((2,-1)\) with slope \(\frac{1}{4}\)

Step-by-Step Solution

Answer

The equation in standard form is \( x - 4y = 6 \).

Step 1: Recall the Point-Slope Form

The point-slope form of a line's equation is given by \( y - y_1 = m(x - x_1) \), where \(m\) is the slope and \((x_1, y_1)\) is a point on the line. In our problem, \( m = \frac{1}{4} \) and the point \((x_1, y_1) = (2, -1)\).

Step 2: Substitute Into Point-Slope Form

Substitute the given slope and point into the point-slope form: \( y - (-1) = \frac{1}{4}(x - 2) \). Simplify this to get \( y + 1 = \frac{1}{4}(x - 2) \).

Step 3: Distribute the Slope

Distribute \( \frac{1}{4} \) within the parenthesis: \( y + 1 = \frac{1}{4}x - \frac{1}{2} \).

Step 4: Rearrange to Slope-Intercept Form

Subtract 1 from both sides: \( y = \frac{1}{4}x - \frac{1}{2} - 1 \). This simplifies to \( y = \frac{1}{4}x - \frac{3}{2} \).

Step 5: Convert to Standard Form

The standard form of a line is \( Ax + By = C \), where A, B, and C are integers, and A should be positive. Start by eliminating the fraction by multiplying the entire equation by 4: \( 4y = x - 6 \). Rearrange to \( -x + 4y = -6 \).

Step 6: Adjust to Standard Form with Positive A

Multiply the entire equation by -1 to make the coefficient of \( x \) positive: \( x - 4y = 6 \).

Key Concepts

Point-Slope Form

The point-slope form of a line is a very handy way to write the equation of a line when we know a specific point on the line and its slope. The general formula for this form is \( y - y_1 = m(x - x_1) \), where:

• \( y_1 \) and \( x_1 \) are the coordinates of the known point on the line \((x_1, y_1)\).
• \( m \) is the slope of the line.

This format is particularly useful because it directly incorporates the slope and a point, making it simple to use. For example, if the slope \( m \) is \( \frac{1}{4} \) and the point is \((2, -1)\), substituting into the formula gives us: \( y - (-1) = \frac{1}{4}(x - 2) \), which simplifies to \( y + 1 = \frac{1}{4}(x - 2) \).
From here, we can solve for \( y \), making it easy to convert this equation into other forms if needed.

Slope-Intercept Form

Once we have the point-slope form, transforming it into the slope-intercept form \( y = mx + b \) is often straightforward. This form clearly shows the slope \( m \) and the y-intercept \( b \). From our example, after simplifying back to \( y + 1 = \frac{1}{4}x - \frac{1}{2} \), we can isolate \( y \) by subtracting 1 from both sides of the equation.

This yields \( y = \frac{1}{4}x - \frac{3}{2} \). In this format:

• \( m \) is still \( \frac{1}{4} \), the slope of the line.
• \( b \), the y-intercept, is \( -\frac{3}{2} \).

Knowing the slope and y-intercept in this form is valuable for graphing, as it instantly shows the vertical position of the line when \( x = 0 \) and how steep the line is.

Standard Form

The standard form of an equation, \( Ax + By = C \), where \( A \), \( B \), and \( C \) must be integers and \( A \) should be a positive number, is quite different from the point-slope or slope-intercept forms. While it is less immediately informative for graphing compared to the slope-intercept form, it is very versatile, especially in algebraic manipulations and solving systems of equations.

Using our equation \( y = \frac{1}{4}x - \frac{3}{2} \) from the slope-intercept form, we aim to clear the fractions by multiplying through by 4, which gives \( 4y = x - 6 \). Arranging this to fit the standard structure, we can rewrite it as \(-x + 4y = -6 \). Finally, by multiplying through by -1, we ensure \( A \) is positive: \( x - 4y = 6 \). In this form, all terms are integers, making it perfectly suited for different types of calculations.