Problem 28 - Algebra 1: Concepts and Skills

Question

Write in standard form an equation of the line that passes through the given point and has the given slope. \((5,-8), m=\frac{1}{2}\)

Step-by-Step Solution

Answer

The equation of the line is \(x - 2y = 21\).

Step 1: Use the Point-Slope Form

Firstly, use the point-slope form of the line equation, \(y - y_{1} = m(x - x_{1})\), where \(x_{1}\) = 5, \(y_{1}\) = -8 and the slope \(m = \frac{1}{2}\). Substituting these values into the point-slope form equation yields: \(y - (-8) = \frac{1}{2}(x - 5)\). Simplifying, this gives \(y + 8 = \frac{1}{2}x - \frac{5}{2}\).

Step 2: Convert to Standard Form

Next, convert this equation to standard form, which is Ax + By = C. To do this, first clear any fractions by multiplying every term by 2. This results in \(2y + 16 = x - 5\). Then arrange the equation to have x and y on the left side and constant on the right: \(x - 2y = 21\).

Step 3: Finalize the Standard Form

Lastly, check to make sure the equation is in the correct standard form. The coefficients A, B, and C should be integers, and the coefficient of x, A, should be positive. Our equation fits this criteria, so the standard form of the given line is \(x - 2y = 21\).

Key Concepts

Point-Slope Form

The point-slope form is a convenient way to write the equation of a line when you know a point on the line and its slope. The formula is written as:

• \(y - y_{1} = m(x - x_{1})\)

where \((x_{1}, y_{1})\) is a specific point on the line and \(m\) represents the slope. This form allows you to see how the line behaves at a particular point.
To solve the exercise using the point-slope form, you substitute \(x_{1} = 5\), \(y_{1} = -8\), and \(m = \frac{1}{2}\) into the formula, resulting in:

• \(y - (-8) = \frac{1}{2}(x - 5)\).

After simplifying, you find:

• \(y + 8 = \frac{1}{2}x - \frac{5}{2}\).

This step allows you to see the impact of the slope and the chosen point directly on the equation.

Slope-Intercept Form

Understanding the slope-intercept form can simplify dealing with linear equations since this form presents the slope and the y-intercept directly. It is expressed as:

• \(y = mx + b\)

where \(m\) is the slope, and \(b\) is the y-intercept, the point where the line crosses the y-axis.
This form is particularly helpful for quickly sketching graphs.
Although the exercise directly converts the point-slope form to the standard form, observing the transformation to the slope-intercept form in the middle is helpful. From \(y + 8 = \frac{1}{2}x - \frac{5}{2}\), you can isolate \(y\) to see:

• \(y = \frac{1}{2}x - \frac{5}{2} - 8\)
• \(y = \frac{1}{2}x - \frac{21}{2}\)

Thus, in slope-intercept form, the slope is \(\frac{1}{2}\), and the y-intercept is \(-\frac{21}{2}\). This provides insight into both the line's steepness and starting point on the graph.

Line Equation

The line equation can take various forms to provide different insights into linear relationships. One of the main equations is the standard form, written as:

• \(Ax + By = C\)

where \(A\), \(B\), and \(C\) are integers, and \(A\) should be positive. This form is particularly useful for finding the intersection of two lines or determining exact values of \(x\) and \(y\).
In our original problem, starting from the point-slope form, you simplify and rearrange:

• Clear fractions by multiplying through by 2: \(2y + 16 = x - 5\).
• Then, re-arrange to get: \(x - 2y = 21\).

Once rearranged, you check that \(A = 1\), \(B = -2\), and \(C = 21\) are all integers. The equation now cleanly describes the line in standard form, ensuring a comprehensive understanding of how it behaves across the coordinate grid.