Problem 35 - Precalculus Mathematics for Calculus

Question

Find an equation of the line that satisfies the given conditions. Through \((-1,-2) ; \quad\) perpendicular to the line \(2 x+5 y+8=0\)

Step-by-Step Solution

Answer

The equation is \( y = \frac{5}{2}x + \frac{1}{2} \).

Step 1: Identify the Slope of the Given Line

First, we need to find the slope of the line given by the equation \(2x + 5y + 8 = 0\). To do this, we rearrange the equation into slope-intercept form \(y = mx + b\). Solve for \(y\): \[ 5y = -2x - 8 \]\[ y = -\frac{2}{5}x - \frac{8}{5} \]So, the slope \(m\) of the given line is \(-\frac{2}{5}\).

Step 2: Determine the Slope of the Perpendicular Line

The slope of a line perpendicular to another is the negative reciprocal of its slope. Thus, the slope \(m_perp\) for the line perpendicular to our given line is: \[ m_{perp} = -\left(-\frac{1}{\frac{2}{5}}\right) = \frac{5}{2} \]

Step 3: Use the Point-Slope Form to Find the Equation

Use the point-slope form of the equation of a line: \[ y - y_1 = m(x - x_1) \]Where \((x_1, y_1)\) is the point \((-1, -2)\) and \(m\) is the slope \(\frac{5}{2}\). Substitute the values: \[ y - (-2) = \frac{5}{2}(x - (-1)) \]

Step 4: Simplify the Equation

First, simplify the expression:\[ y + 2 = \frac{5}{2}(x + 1) \]Distribute \(\frac{5}{2}\):\[ y + 2 = \frac{5}{2}x + \frac{5}{2} \]Subtract 2 from both sides to solve for \(y\):\[ y = \frac{5}{2}x + \frac{5}{2} - 2 \]Convert \(-2\) to have a common denominator:\[ y = \frac{5}{2}x + \frac{5}{2} - \frac{4}{2} \]Combine the fractions:\[ y = \frac{5}{2}x + \frac{1}{2} \]

Step 5: Write the Final Equation

The final equation of the line passing through \((-1, -2)\) and perpendicular to the line \(2x + 5y + 8 = 0\) is:\[ y = \frac{5}{2}x + \frac{1}{2} \]

Key Concepts

Perpendicular Lines

When two lines are perpendicular, they intersect at a right angle (90 degrees). This means that the angles formed where they meet are right angles. A distinctive and important characteristic of perpendicular lines is their slopes. If the slope of one line is known, the slope of the line perpendicular to it is the negative reciprocal. This concept can be explained more simply:

• The slope of the original line, denoted as \( m \), is the coefficient of \( x \) when the line is expressed in the slope-intercept form \( y = mx + b \).
• For a line perpendicular to another, its slope, denoted as \( m_{\text{perpendicular}} \), is the negative reciprocal. Mathematically, if \( m = a/b \), then \( m_{\text{perpendicular}} = -b/a \).

For example, if you have a line with a slope of \(-\frac{2}{5}\), like in the given exercise, the slope of the line that is perpendicular to it will be \(\frac{5}{2}\), flipping the fraction and changing the sign. This basic principle is crucial when working with lines, especially in geometry and algebra.

Slope-Intercept Form

The slope-intercept form of a line’s equation is an intuitive way to understand the relationship between the slope of the line and its y-intercept. This form is expressed as \( y = mx + b \), where:

• \( m \) represents the slope of the line, which indicates the steepness or inclination. It tells us how much \( y \) changes for a unit change in \( x \).
• \( b \) is the y-intercept, the point where the line crosses the y-axis. It is the value of \( y \) when \( x = 0 \).

Using the slope-intercept form is convenient for graphing because it gives a clear picture of both the slope and where the line will intersect the y-axis. In our exercise, once we found the slope of the perpendicular line, we used it to write the equation in this form, making it easier to visualize and work with: \( y = \frac{5}{2}x + \frac{1}{2} \). This equation illustrates that the line rises 2.5 units vertically for every unit it moves horizontally, and it meets the y-axis at \( \frac{1}{2} \).

Point-Slope Form

The point-slope form of a line is particularly useful when you know a point on the line and the slope. This form is written as \( y - y_1 = m(x - x_1) \), where:

• \((x_1, y_1)\) is a specific point through which the line passes.
• \( m \) is the slope of the line.

To derive the equation for a line when given a point and the slope, you simply plug these values into the formula. For instance, in our exercise, we used the point \((-1, -2)\) and the slope \(\frac{5}{2}\) to find the line’s equation. Plugging these into the point-slope form gives \( y + 2 = \frac{5}{2}(x + 1) \).
This is then simplified into the slope-intercept form of the equation, making it easy to interpret. The point-slope form is an essential tool as it provides a straightforward method to find the equation of a line when a point and a slope are known, bridging the gap between raw information and practical expressions in mathematics.