Problem 59 - Elementary Algebra

Question

Contains the origin and is perpendicular to the line \(-2 x+3 y=8\)

Step-by-Step Solution

Answer

Equation: \(y = -\frac{3}{2}x\).

Step 1: Identify the Slope of the Given Line

The given equation of the line is \(-2x + 3y = 8\). To find the slope, rewrite the equation in slope-intercept form, \(y = mx + b\). Start by isolating \(y\):\[ 3y = 2x + 8 \]Divide every term by 3:\[ y = \frac{2}{3}x + \frac{8}{3} \]The slope \(m\) of the given line is \(\frac{2}{3}\).

Step 2: Find the Slope of the Perpendicular Line

Lines that are perpendicular have slopes that multiply to -1. Thus, if one line has slope \(\frac{2}{3}\), the perpendicular line's slope \(m'\) satisfies:\[ \frac{2}{3} \times m' = -1 \]Solving for \(m'\), we have:\[ m' = -\frac{3}{2} \]

Step 3: Write the Equation of the Perpendicular Line Through the Origin

The perpendicular line passes through the origin \((0, 0)\) and has the slope \(-\frac{3}{2}\). The point-slope form of a line's equation is \(y - y_1 = m(x - x_1)\). Substituting \(m = -\frac{3}{2}\), \(x_1 = 0\), and \(y_1 = 0\):\[ y - 0 = -\frac{3}{2}(x - 0) \]This simplifies to:\[ y = -\frac{3}{2}x \]

Step 4: Verify the Equation Satisfies Both Conditions

Verify that the line \(y = -\frac{3}{2}x\) satisfies both required conditions:1. **Contains the Origin**: Plug in \(x = 0\), \(y = 0\): \(0 = -\frac{3}{2} \times 0\), which is true.2. **Perpendicular to the Given Line**: The slope \(-\frac{3}{2}\) was calculated to be the negative reciprocal of \(\frac{2}{3}\), confirming perpendicularity.

Key Concepts

Slope-Intercept Form

The slope-intercept form is a way to express a linear equation so it's easy to identify both the slope and the y-intercept. It is written as:\[ y = mx + b \]Where:- \( m \) represents the slope of the line.- \( b \) is the y-intercept, the point where the line crosses the y-axis.Converting an equation into the slope-intercept form helps in easily reading the slope, which indicates how steep the line is. For example, in the original exercise, we started with the equation \(-2x + 3y = 8\), which is not in the slope-intercept form. By isolating \( y \), we convert it into:\[ y = \frac{2}{3}x + \frac{8}{3} \]This tells us that for every unit increase in \( x \), \( y \) increases by \( \frac{2}{3} \). The y-intercept here is \( \frac{8}{3} \), meaning the line crosses the y-axis at the point (0, \( \frac{8}{3} \)). The slope-intercept form simplifies graphing the line and identifying its characteristics.

Point-Slope Form

The point-slope form is particularly useful for writing the equation of a line when you know a point on the line and the slope. This form is given by:\[ y - y_1 = m(x - x_1) \]Where:- \( m \) is the slope.- \( (x_1, y_1) \) is a specific point on the line.This format is flexible and valuable, especially when dealing with perpendicular lines. If a line is perpendicular to another, and its slope is known, you can use the point-slope form to write the new line’s equation quickly. From the exercise, the perpendicular line passes through the origin \[ (0, 0) \] and has a slope of \(-\frac{3}{2}\). Using the point-slope form:\[ y - 0 = -\frac{3}{2}(x - 0) \]This simplifies to:\[ y = -\frac{3}{2}x \]This shows how straightforward it is to determine the equation of the line directly from the slope and a known point.

Negative Reciprocal

A negative reciprocal is a fundamental concept when dealing with perpendicular lines. Two lines are perpendicular if the product of their slopes equals -1. Knowing this, you can quickly figure out the slope of a perpendicular line.To find the negative reciprocal:- Take the original slope.- Invert the fraction (flip the numerator and denominator).- Change the sign.In the example problem, the original line's slope is \( \frac{2}{3} \). Its negative reciprocal is thus:1. Invert: \( \frac{3}{2} \).2. Change the sign: \(-\frac{3}{2} \).Thus, the perpendicular line will have a slope of \(-\frac{3}{2}\), ensuring the two lines meet at a right angle. Understanding negative reciprocals is crucial when working on geometry problems involving perpendicularity, as it allows you to systematically derive solutions rather than relying on trial and error.