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Chapter 2: Problem 45

Write an equation (a) in standard form and (b) in slope-intercept form for the line described through \((-1,4),\) parallel to \(x+3 y=5\)

Short Answer

Expert verified
Standard form: x + 3y = 11. Slope-intercept form: y = \(-\frac{1}{3}\)x + \(\frac{11}{3}\).

Step by step solution

01

Identify the slope of the given line

Start by putting the given equation in slope-intercept form, which is: x + 3y = 5. Subtract x from both sides: 3y = -x + 5. Divide by 3: y = \(-\frac{1}{3}\)x + \(\frac{5}{3}\). The slope-intercept form is y = mx + b, so the slope (m) of the given line is \(-\frac{1}{3}\)
02

Use the slope for the new line

Since parallel lines have the same slope, the slope of our new line through (-1, 4) will also be \(-\frac{1}{3}\).
03

Use the point-slope form to write the equation

The point-slope form is: y - y_1 = m(x - x_1). Substituting in the given point (-1,4) and the slope \(-\frac{1}{3}\): y - 4 = \(-\frac{1}{3}\)(x + 1). y -4 = \(-\frac{1}{3}\)x - \(\frac{1}{3}\)
04

Convert to slope-intercept form

From the equation y - 4 = \(-\frac{1}{3}\)x - \(\frac{1}{3}\), add 4 to both sides to isolate y: y = \(-\frac{1}{3}\)x - \(\frac{1}{3}\) + 4 y = \(-\frac{1}{3}\)x + \(\frac{11}{3}\)
05

Convert to standard form

Starting from the slope-intercept form: y = \(-\frac{1}{3}\)x + \(\frac{11}{3}\), multiply every term by 3 to eliminate the fraction: 3y = -x + 11 Rearrange to get the standard form: x + 3y = 11

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

slope-intercept form
Understanding the slope-intercept form of a linear equation is crucial in coordinate geometry. The slope-intercept form is written as \(y = mx + b\), where \(m\) represents the slope of the line and \(b\) is the y-intercept, the point where the line crosses the y-axis. This form is particularly useful for quickly identifying the slope and y-intercept.
For example, in the exercise, we transformed the given equation \(x + 3y = 5\) into slope-intercept form by isolating \(y\): \(y = -\frac{1}{3}x + \frac{5}{3}\). From this, we can see the slope (\(m\)) is \(-\frac{1}{3}\), and the y-intercept (\(b\)) is \(\frac{5}{3}\).
standard form
In addition to the slope-intercept form, linear equations can also be expressed in standard form. The standard form of a linear equation is \(Ax + By = C\), where \(A\), \(B\), and \(C\) are integers, and \(A\) should be non-negative.
For example, in the exercise, we took the slope-intercept form \(y = -\frac{1}{3}x + \frac{11}{3}\) and converted it to standard form by eliminating the fractions. We multiplied every term by 3, resulting in \(-x + 3y = 11\), and then rearranged it to \(x + 3y = 11\).
parallel lines
Parallel lines are a key concept in geometry. Two lines are parallel if they have the same slope but different y-intercepts. This means they never intersect.
In the given exercise, since the line we were working with needed to be parallel to the line given by \(x + 3y = 5\), we ensured that our new line had the same slope, \(-\frac{1}{3}\). This consistency in slope ensures that the two lines remain parallel, regardless of their specific y-intercepts.
point-slope form
The point-slope form is another way to write the equation of a line. It is useful when you have a specific point and a slope. The point-slope form is written as \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \( (x_1, y_1) \) is a point on the line.
In the exercise, we used the point \((-1, 4)\) and the slope \(-\frac{1}{3}\) to write the equation: \( y - 4 = -\frac{1}{3}(x + 1)\). This form simplifies the process of finding the line passing through a given point with a specific slope.
coordinate geometry
Coordinate geometry is a branch of mathematics used to define and represent geometric shapes in a numerical way. It relies heavily on the coordinate plane, consisting of an x-axis and a y-axis.
For example, in the exercise, we utilized coordinate geometry concepts to graph and understand the equations of lines. By converting between different forms (slope-intercept, standard, and point-slope) and leveraging the properties of parallel lines, we could precisely describe the line passing through \((-1, 4)\) parallel to \(x + 3y = 5\) on the coordinate plane.

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