Problem 56 - Algebra and Trigonometry Real Mathematics, Real People

Question

Write an equation of the line that passes through the points. Use the slope- intercept form (if possible). If not possible, explain why and use the general form. Use a graphing utility to graph the line (if possible). $$(1,1),\left(6,-\frac{2}{3}\right)$$

Step-by-Step Solution

Answer

The equation of the line is \(y=(-1/3)x + 4/3\).

Step 1: Calculate the slope

Use the formula for the slope: \(m=(y_2-y_1)/(x_2-x_1)\). Substituting the given points \((1,1)\) and \((6,-2/3)\) into this formula gives \(m=(-2/3-1)/(6-1) = -5/15=-1/3\).

Step 2: Find the y-intercept

Substitute the coordinates of one of the given points and the slope into the slope-intercept form equation to solve for c. Using the point \((1,1)\) gives \(1=(-1/3)*1 + c\). Solving for c gives \(c=1+(1/3)=4/3\).

Step 3: Construct the equation of the line

Substitute the slope and y-intercept into the slope-intercept form equation. This gives \(y=(-1/3)x + 4/3\).

Key Concepts

Linear Equations

Linear equations are fundamental mathematical expressions that describe a straight line on a graph. To put it simply, these equations have variables that have a maximum exponent of one.
This characteristic results in a graph represented by a straight, uniform line. The most common form of a linear equation is the slope-intercept form, expressed as \(y = mx + c\). Here,

• \(y\) is the dependent variable or the output of the equation.
• \(x\) is the independent variable or input.
• \(m\) is the slope of the line, indicating its steepness.
• \(c\) is the y-intercept, representing the point at which the line intersects the y-axis.

Linear equations are everywhere in algebra and calculus. They are crucial for modeling and understanding linear relationships between two variables.

Graphing Lines

When you graph a linear equation, you create a visual representation of the relationship expressed by that equation. The primary aim is to plot a straight line that showcases the equation's dynamics.
Start with the slope-intercept form \(y = mx + c\), as it's convenient for graphing.

• Begin by locating the y-intercept \(c\) on the y-axis. It's your starting point.
• Next, utilize the slope \(m\). The slope \(-1/3\), for instance, means for every 3 units you move right horizontally, you move 1 unit down vertically.
• Use these points to plot your line confidently across the graph.

This visual approach helps in understanding the behavior of the line. Whether it slopes upwards or downwards is determined by the sign and value of \(m\). A positive slope indicates an upward trend, while a negative slope reveals a downward slant.

Slope Calculation

One of the first steps in describing a linear equation is calculating its slope, often symbolized by \(m\). The slope quantifies how sharply the line tilts, indicating a direction and rate of increase or decrease.
To calculate the slope from two given points \((x_1, y_1)\) and \((x_2, y_2)\), use the formula:\[m = \frac{y_2 - y_1}{x_2 - x_1}\]For the points \((1,1)\) and \((6, -\frac{2}{3})\), the calculation becomes \(\frac{-\frac{2}{3} - 1}{6 - 1} = \frac{-\frac{5}{3}}{5} = -\frac{1}{3}\).

• A slope of -1/3 reveals that for every 3 units moved to the right, the line moves down 1 unit.
• The line trends downward because the slope is negative.

Understanding how to compute and interpret the slope is crucial when working with linear equations, as it defines the line's orientation on the graph.

Y-intercept Finding

Finding the y-intercept \(c\) is a key step in writing the equation of a line. The y-intercept is the point where the line crosses the y-axis, meaning \(x\) is zero here.
Once you have the slope \(m\), you can find \(c\) using the slope-intercept form \(y = mx + c\).
Let's insert one of our points \((1, 1)\) and the known slope \(-\frac{1}{3}\) into the equation:\[1 = -\frac{1}{3} \cdot 1 + c\]Solving for \(c\) involves:

• Multiply: \(-\frac{1}{3}\times 1 = -\frac{1}{3}\)
• Rearrange: \(1 = -\frac{1}{3} + c\)
• Add \(\frac{1}{3}\) to both sides: \(c = 1 + \frac{1}{3} = \frac{4}{3}\)

Thus, \(c = \frac{4}{3}\) confirms that the line crosses the y-axis at this point. Calculating \(c\) ensures the equation of the line accurately reflects its position on the graph.