Problem 36

Question

Write an equation in slope-intercept form of the line satisfying the given conditions. The line passes through \((2,6)\) and has the same \(y\) -intercept as the line whose equation is \(x-3 y=18.\)

Step-by-Step Solution

Verified
Answer
The equation of the line is \(y = 6x - 6\).
1Step 1: Finding the y-intercept of given line
Rewrite the given equation \(x-3y = 18\) in slope-intercept form to find the y-intercept. Solving for \(y\), we have \(3y = x - 18\), hence \(y = \frac{1}{3}x - 6\). The y-intercept is -6.
2Step 2: Calculating the slope of the new line
We have the point \((2,6)\) and the y-intercept \((0,-6)\). So we can find the slope (m) using the formula: \(m=\frac{y2 - y1}{x2 - x1}\). Substituting in the given points, we get \(m=\frac{6 - (-6)}{2 - 0} = 6\).
3Step 3: Write the equation of the line in slope-intercept form
Given that the slope (m) is 6 and the y-intercept (c) is -6, we substitute these values into the slope-intercept form \(y = mx + c\) to get \(y = 6x - 6\). This is the equation of the line.

Key Concepts

Writing Linear EquationsY-interceptSlope CalculationAlgebraic Manipulation
Writing Linear Equations
Understanding how to write linear equations is a crucial skill in algebra. A linear equation forms a straight line when graphed, and the most common way to write it is in the slope-intercept form, which is expressed as
\( y = mx + b \), where
m represents the slope of the line, and b is the y-intercept. The y-intercept is the point where the line crosses the y-axis, and the slope indicates the steepness of the line.
When you're given a point through which the line passes and another equation to refer to, as in the original exercise, the task is to find both values, m and b, to completely determine the desired line's equation. Once you have both the slope and the y-intercept, the process of writing the equation is straightforward. Substitute these values into the slope-intercept form, and you've got the required linear equation.
Y-intercept
The y-intercept is a fundamental concept when working with linear equations. It's the point where the line crosses the y-axis, and in the equation
\( y = mx + b \), it's represented by b. In our exercise, the y-intercept is determined by manipulating the given secondary equation,
\( x - 3y = 18 \), into slope-intercept form, resulting in
\( y = \frac{1}{3}x - 6 \).The number -6 in this rearranged form is our y-intercept for the line we're seeking. This value is crucial because it doesn't change regardless of where on the line you're looking, making it a reliable starting point for constructing your line equation or for graphing purposes.
Slope Calculation
The slope calculation is an essential step in forming the equation of a line. The slope, typically represented as m, shows how much the y-value of a point on a line changes for a one-unit increase in the x-value. Mathematically, it is computed using the formula
\( m = \frac{y_2 - y_1}{x_2 - x_1} \).In the exercise, we use the slope calculation to find how steep our line will be. We already have one point,
\( (2,6) \), and the y-intercept at
\( (0,-6) \), giving us a second point. Plugging these points into the formula gives us a slope of
\( m = \frac{6 - (-6)}{2 - 0} = 6 \). This value indicates that for every one unit you move to the right along the x-axis, the y-value increases by six units—a pretty steep line!
Algebraic Manipulation
The term algebraic manipulation encompasses a range of techniques used to rearrange and solve equations. In our context, we apply these techniques to rewrite the given equation in a form that helps us identify the slope and y-intercept.
The given equation was
\( x - 3y = 18 \), which we manipulate to isolate y by dividing all terms by -3. The result is
\( y = \frac{1}{3}x - 6 \), an equation where it's easy to spot the slope and y-intercept. Understanding algebraic manipulation is crucial for solving for these values, which are then used to write the final linear equation in slope-intercept form. It’s important to feel comfortable with operations such as addition, subtraction, multiplication, division, and the use of inverse operations to solve for variables to succeed in algebra.