Problem 46 - A Graphical Approach to College Algebra

Question

Find the equation of the line satisfying the given conditions, giving it in slope-intercept form if possible. Through \((-2,0),\) perpendicular to \(8 x-3 y=7\)

Step-by-Step Solution

Answer

The equation is \(y = -\frac{3}{8}x - \frac{3}{4}\).

Step 1: Identify the slope of the given line

The given line is in the form \(8x - 3y = 7\). To find the slope, we convert this to the slope-intercept form \(y = mx + b\), where \(m\) is the slope. Start by solving for \(y\) in terms of \(x\): \[ -3y = -8x + 7 \] Now divide every term by \(-3\): \[ y = \frac{8}{3}x - \frac{7}{3} \] Therefore, the slope of the given line is \(m = \frac{8}{3}\).

Step 2: Find the slope of the perpendicular line

Lines that are perpendicular have slopes that are negative reciprocals. The slope of the imaginary perpendicular line will be \(-\frac{1}{m}\), where \(m = \frac{8}{3}\). Thus, the slope of the perpendicular line is \(-\frac{3}{8}\).

Step 3: Use point-slope form with the known point

The point given is \((-2, 0)\), and the slope of the perpendicular line is \(-\frac{3}{8}\). Plug these into the point-slope form equation: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (-2, 0)\). Substitute \(m = -\frac{3}{8}\): \[ y - 0 = -\frac{3}{8}(x - (-2)) \] Which simplifies to: \[ y = -\frac{3}{8}(x + 2) \]

Step 4: Convert to slope-intercept form

The equation from the previous step is in point-slope form. We need to convert it to slope-intercept form \(y = mx + b\). Expand the equation: \[ y = -\frac{3}{8}x - \frac{3}{8} \times 2 \] Simplify: \[ y = -\frac{3}{8}x - \frac{3 \times 2}{8} \] Further simplification leads to: \[ y = -\frac{3}{8}x - \frac{3}{4} \] Hence, the slope-intercept form of the equation is \(y = -\frac{3}{8}x - \frac{3}{4}\).

Key Concepts

Equation of a Line

Understanding the equation of a line is vital for working with linear equations. When talking about the equation of a line, the most common form you may encounter is the slope-intercept form. It is expressed as \( y = mx + b \), where:

• \( y \) is the dependent variable (usually representing vertical position),
• \( x \) is the independent variable (usually representing horizontal position),
• \( m \) is the slope of the line, which indicates the steepness and direction,
• \( b \) is the y-intercept, which is the point where the line crosses the y-axis.

The slope \( m \) tells us how much \( y \) changes for a unit change in \( x \). If \( m \) is positive, the line slopes upwards as it moves from left to right. If \( m \) is negative, it slopes downwards. The value of the y-intercept \( b \) gives us a starting point on the y-axis. Understanding and converting to slope-intercept form is essential when graphing a line or comparing different linear equations.

Perpendicular Lines

Perpendicular lines are two lines that intersect at a right angle, specifically, 90 degrees. In coordinate geometry, when working with equations of lines, perpendicularity is represented mathematically through the slopes of the lines. Two lines are perpendicular if the product of their slopes is \(-1\). In simpler terms, their slopes are negative reciprocals of each other. For instance, if one line has a slope \( \frac{8}{3} \), a line perpendicular to it will have a slope \( -\frac{3}{8} \).
Negative reciprocals are found by flipping the fraction and changing the sign. Understanding this relationship helps in solving problems where you need to find equations of lines that are perpendicular to a given line. This concept is a key component in geometry and can assist in identifying right angles within various figures on a coordinate plane.

Point-Slope Form

The point-slope form is another way to express the equation of a line. It is particularly useful when you know the slope of the line and a point on the line. The formula is given by \( y - y_1 = m(x - x_1) \), where:

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

Using the point-slope form is advantageous for quickly finding the equation of a line when a direct slope and a point are provided. This form is also a bridge between different expressions of linear equations. Once you have the point-slope form, you can easily rearrange it into slope-intercept form \(y = mx + b\) by expanding the terms. By learning how to manipulate these forms, one can seamlessly move between different scenarios in math, making the point-slope method a handy tool in the algebra toolkit.