Problem 28 - Elementary Algebra

Question

You are given one point on a line and the slope of the line. Find the coordinates of three other points on the line. $$(-6,-2), m=\frac{2}{5}$$

Step-by-Step Solution

Answer

Three points on the line are (0, 2/5), (5, 12/5), and (10, 22/5).

Step 1: Understanding the Given Information

We are given a point on the line, \((-6,-2)\), and the slope of the line, \(m=\frac{2}{5}\). The slope indicates that for every 5 units you move horizontally to the right, you move 2 units vertically up. This information will help us find other points on the line.

Step 2: Finding the Slope Intercept Form

First, we need to write the equation of the line in the slope-intercept form, which is \(y = mx + b\). We know the slope \(m = \frac{2}{5}\) and one point \((-6,-2)\), so we can substitute these into the equation to find \(b\).

Step 3: Substituting into the Equation

Using the known point \((-6,-2)\) in the equation \(y = mx + b\), substitute \(x = -6\), \(y = -2\), and \(m = \frac{2}{5}\): \(-2 = \frac{2}{5}(-6) + b\).

Step 4: Solving for the Intercept

Calculate the value of \(b\):\(-2 = -\frac{12}{5} + b\).Convert \(-2\) to a fraction, \(-\frac{10}{5}\), and solve for \(b\):\(-\frac{10}{5} + \frac{12}{5} = b\), so, \(b = \frac{2}{5}\).

Step 5: Writing the Equation of the Line

Now that we have both the slope and the y-intercept, the equation of the line is:\(y = \frac{2}{5}x + \frac{2}{5}\).

Step 6: Selecting Different X Values

To find three other points, we can select different values for \(x\), substitute each into the equation, and find the corresponding \(y\) value.

Step 7: Substitute X = 0

If \(x = 0\):\(y = \frac{2}{5}(0) + \frac{2}{5} = \frac{2}{5}\)The point is \((0, \frac{2}{5})\).

Step 8: Substitute X = 5

If \(x = 5\):\(y = \frac{2}{5}(5) + \frac{2}{5} = 2 + \frac{2}{5} = \frac{12}{5}\)The point is \((5, \frac{12}{5})\).

Step 9: Substitute X = 10

If \(x = 10\):\(y = \frac{2}{5}(10) + \frac{2}{5} = 4 + \frac{2}{5} =\frac{22}{5}\)The point is \((10, \frac{22}{5})\).

Step 10: Listing the Points

Coordinates of three additional points on the line are:- \((0, \frac{2}{5})\)- \((5, \frac{12}{5})\)- \((10, \frac{22}{5})\)

Key Concepts

Slope-Intercept Form

The slope-intercept form of a linear equation is a very convenient way to represent a line. It's most commonly expressed as \( y = mx + b \). Here, \( m \) is the slope of the line, and \( b \) is the y-intercept, which is the point where the line crosses the y-axis.

• The slope, \( m \), describes the steepness of the line. If it's positive, the line rises as you move from left to right. If it's negative, the line falls.
• The y-intercept, \( b \), tells you where the line intersects the y-axis, essentially giving you an initial starting point on the graph.

For the problem we are considering, knowing the slope as \( \frac{2}{5} \), we understand that for every 5 units we move to the right on the x-axis, the line moves up by 2 units on the y-axis. This consistent pattern helps us determine other points along the line effectively.

Coordinate Geometry

Coordinate geometry lets us explore lines using algebra. It gives us a way to describe a line using numbers and equations. By using the coordinate plane, every point on a line can be identified using a pair of numbers, usually written as \((x, y)\).When dealing with a linear equation in slope-intercept form, we use the slope \( m \) and y-intercept \( b \) to find the position of the line in the coordinate plane.

• The x-coordinate represents the horizontal position of a point in the plane.
• The y-coordinate represents the vertical position.

This method makes plotting a line simple and precise by using the concrete relationships captured in the equation of the line. For instance, every point \((x, y)\) on the line can be found by solving the equation \( y = mx + b \) for a given \( x \) value.

Finding Points on a Line

To find points on a line given an equation, such as \( y = \frac{2}{5}x + \frac{2}{5} \), we can choose different \( x \)-values and calculate the corresponding \( y \)-values. This equation tells us exactly what \( y \) will be for any \( x \) we choose. For example:

• If \( x = 0 \), substituting into the equation gives \( y = \frac{2}{5}(0) + \frac{2}{5} = \frac{2}{5} \). Therefore, the point \( (0, \frac{2}{5}) \) lies on the line.
• If \( x = 5 \), substituting gives \( y = \frac{2}{5}(5) + \frac{2}{5} = \frac{12}{5} \), making the point \( (5, \frac{12}{5}) \).
• If \( x = 10 \), then \( y = \frac{2}{5}(10) + \frac{2}{5} = \frac{22}{5} \), resulting in the point \( (10, \frac{22}{5}) \).

By repeating this process with other x-values, we can find any number of points on the line, confirming its straight course without any curvature. Thus, this method provides a clear understanding of linear relationships.