Problem 27 - 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 \((2,-3)\) and parallel to $$ x+2 y-4=0 $$

Step-by-Step Solution

Answer

The equation is \(x + 2y + 4 = 0\).

Step 1: Understand the Problem

We need to find the equation of a line passing through the point \((2, -3)\) and parallel to the given line \(x + 2y - 4 = 0\). We will put the equation in standard form.

Step 2: Identify the Slope of the Parallel Line

The given line is in the form \(Ax + By + C = 0\). To find the slope, we convert it to the slope-intercept form, \(y = mx + b\). Solve \(x + 2y - 4 = 0\) to get \(2y = -x + 4\), hence \(y = -\frac{1}{2}x + 2\). The slope \(m\) is \(-\frac{1}{2}\). For parallel lines, the slope is the same.

Step 3: Use Point-Slope Form

Use the point-slope form \(y - y_1 = m(x - x_1)\) where \((x_1, y_1) = (2, -3)\) and \(m = -\frac{1}{2}\). Substitute these values: \(y + 3 = -\frac{1}{2}(x - 2)\).

Step 4: Simplify the Equation

Distribute the \(-\frac{1}{2}\): \(y + 3 = -\frac{1}{2}x + 1\). Now, subtract 3 from both sides: \(y = -\frac{1}{2}x - 2\).

Step 5: Convert to Standard Form

Multiply every term by 2 to eliminate the fraction: \(2y = -x - 4\). Rearrange to get standard form \(Ax + By + C = 0\): add \(x\) to both sides to get \(x + 2y + 4 = 0\). This is the standard form equation.

Key Concepts

Slope-Intercept Form

The slope-intercept form of a linear equation is a widely used way to express an equation of a line. It is given by the formula:
\[ y = mx + b \]
Where:

• \( y \) is the dependent variable or the output of the function.
• \( x \) is the independent variable or the input of the function.
• \( m \) represents the slope of the line, indicating how steep the line is.
• \( b \) is the y-intercept, which is the point where the line crosses the y-axis.

This form is particularly useful when you need to quickly identify the slope and the y-intercept of the line. For example, consider the line equation given as:
\[ y = -\frac{1}{2}x + 2 \]
Here, the slope \( m \) is -\frac{1}{2}, which means the line goes downwards as you move from left to right. The y-intercept is 2, indicating that the line crosses the y-axis at the point (0,2). This form is very convenient for easily graphing lines and understanding how changes in \( x \) affect \( y \).

Standard Form

Standard form of a linear equation is an alternative way of expressing the equation of a line. It is represented as:
\[ Ax + By + C = 0 \]
Where:

• \( A, B, \) and \( C \) are integers, and \( A \) should not be negative.
• \( x, y \) are variables.

This form is particularly convenient when dealing with systems of equations or when you need to rearrange terms. In our exercise, the line equation was given as:
\[ x + 2y - 4 = 0 \]
To convert this into standard form while maintaining the parallel feature, we adjust for any specific point. If we want a line through (2,-3) parallel to this, we will end up with the equation:
\[ x + 2y + 4 = 0 \]
In this exercise, the process involved rearranging the equation and maintaining integer values for the coefficients, which simplifies the analysis and manipulation of algebraic expressions.

Point-Slope Form

The point-slope form is another common way to write the equation of a line. It is useful when you know one point on the line and the slope. The formula is:
\[ y - y_1 = m(x - x_1) \]
Where:

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

This form is especially helpful for writing the equation of a line when you have a single point and the line's slope. In the exercise, the point-slope form was utilized to derive the equation of a line by substituting the known point \((2, -3)\) and the slope \(-\frac{1}{2}\):
\[ y + 3 = -\frac{1}{2}(x - 2) \]
After performing some algebraic manipulation to simplify the expression, it was eventually converted to other forms. Deploying the point-slope formula gives a very direct way to generate a line equation, especially when aiming to express the line's behavior relative to its slope and a particular point, before transforming it into simpler forms like slope-intercept or standard form.