Problem 51 - Algebra 2

Question

CHALLENGE If the graph of the equation \(a x+3 y=9\) is perpendicular to the graph of the equation \(3 x+y=-4,\) find the value of \(a .\)

Step-by-Step Solution

Answer

The value of \(a\) is -1.

Step 1: Identify Slopes in Standard Form

The equation of the line can be rearranged from standard form to slope-intercept form. The general form for a line is \(ax+by=c\), which can be rearranged to \(y=-\frac{a}{b}x + \frac{c}{b}\) to identify the slope. Let's convert the first line, \(3x + y = -4\), to the slope-intercept form. Here, \(b = 1\). The slope of this line \(m_1\) is \(-3\).

Step 2: Understand Perpendicularity Condition

Two lines are perpendicular to each other if the product of their slopes is \(-1\). So, if the slope of the first line \(m_1 = -3\) and the slope of a line perpendicular to it \(m_2\), the condition becomes \(m_1 \times m_2 = -1\), or \(-3 \times m_2 = -1\).

Step 3: Find the Slope of the Second Line

We already know that the line perpendicular to \(3x + y = -4\) has an unknown slope \(m_2\). Using the perpendicularity condition \(-3 \times m_2 = -1\), we solve to find \(m_2\) as follows: \(m_2 = \frac{1}{3}\).

Step 4: Convert Second Line to Slope Form and Find \\(a\\)

For the line \(ax + 3y = 9\), convert to slope-intercept form \(y = -\frac{a}{3}x + 3\). The slope of this line \(m_2 = -\frac{a}{3}\). Using the above found \(m_2 = \frac{1}{3}\), equate them to find \(-\frac{a}{3} = \frac{1}{3}\).

Step 5: Solve for \\(a\\)

Solving \(-\frac{a}{3} = \frac{1}{3}\) gives \(a = -1\) by multiplying each side by \(-3\).

Key Concepts

Slope of a Line

The slope of a line is a numerical measure of its steepness, usually denoted by the letter \(m\). It involves the amount of vertical change for a unit of horizontal change between two points on the line.
For any two points \( (x_1, y_1) \) and \( (x_2, y_2) \) on the line, the slope \(m\) is defined as:\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]The slope of a line tells us how much \(y\) changes per unit change in \(x\).
A positive slope indicates the line ascends from left to right, while a negative slope indicates it descends. A zero slope denotes a horizontal line, and an undefined slope means a vertical line.
In the context of our exercise, when lines are perpendicular, their slopes multiply to give \(-1\). This characteristic helps determine unknown slopes when one is known.

Standard Form of a Line

The standard form of a line is expressed as \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) are variables representing coordinates.
This form is handy for identifying intercepts and is often used in algebraic calculations.
You can rearrange this form into different forms to identify the slope, like converting to the slope-intercept form. For instance, the line \(3x + y = -4\) can be rewritten as \(-3x + y = -4\), and further as:\[ y = -3x - 4 \] This conversion unveils the slope as \(-3\). Being adept at switching between line forms can simplify solving problems involving perpendicularity and slope calculations.

Slope-Intercept Form

The slope-intercept form of a linear equation is \(y = mx + b\), where \(m\) is the slope and \(b\) is the \(y\)-intercept (the point where the line crosses the \(y\)-axis).
It is a highly useful form for quick graphing of straight lines and understanding the line's direction and position instantly.- This form explicitly shows the rate of change and the initial value when \(x = 0\).Converting from standard form, as seen in the earlier line equation \(ax + 3y = 9\) becomes:\[ y = -\frac{a}{3}x + 3 \]Knowing how to transition between these forms allows us to identify the slope and subsequently solve for problems related to perpendicularity, as we saw by equating to \(\frac{1}{3}\) to find \(a\).

Linear Equations

Linear equations describe straight lines in algebra and can be written in various forms, including the standard form and slope-intercept form.
These equations take the shape \(ax + by + c = 0\) and graph into lines.- Linear equations appear frequently in both algebra and geometry problems.- Recognizing the different forms, converting between them, and interpreting their components is vital for tackling complex algebraic equations.
In our exercise, we dealt with two linear equations in standard form, converting them to slope-intercept form to solve for unknown variables and determine perpendicular relationship conditions. Understanding linear equations' versatility and capability is key to mastering such exercises.