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Chapter 3: Problem 56

Find an equation of the line passing through the given points. $$(1,4),(3,4)$$

Short Answer

Expert verified
The equation of the line passing through the points (1,4) and (3,4) is \(y = 4\).

Step by step solution

01

Calculate the Slope

The slope (m) is calculated by the formula: \(m = \frac{{y_2 - y_1}}{{x_2 - x_1}}\)\nIn this case, (1,4) is (x1,y1) and (3,4) is (x2,y2).\nSo, \(m = \frac{{4 - 4}}{{3 - 1}} = 0.\)
02

Substituting the Slope and the Point into the Equation

The standard equation for a line is \(y = mx + b\), where m stands for the slope and b for the y-intercept. Substitute the value of m and the coordinates of one point into the equation. Taking the point (1,4): \(4 = 0*1 + b.\) Solving for b gives \(b = 4\).
03

Writing the Equation of the Line

By substituting the values of slope (m) and y-intercept (b) into the equation, the equation of the line becomes \(y = 0x + 4\) or simply \(y = 4\).

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Key Concepts

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

Slope Calculation
When trying to find the slope of a line, you are determining how steep the line is. It's like asking how much the line rises or falls as it moves horizontally. The formula for slope \(m\) is: \(m = \frac{{y_2 - y_1}}{{x_2 - x_1}}\). This formula calculates the change in the \(y\)-coordinates (vertical change) over the change in the \(x\)-coordinates (horizontal change). Let's break it down:
  • \(y_2\) and \(y_1\) represent two \(y\)-values from the different points on the line.
  • \(x_2\) and \(x_1\) are the corresponding \(x\)-values for these points.
For the points \((1, 4)\) and \((3, 4)\), the slope calculation would be: \[m = \frac{{4 - 4}}{{3 - 1}} = \frac{0}{2} = 0\] This result of zero means the line is perfectly horizontal, indicating there is no vertical change as you move along the line horizontally.
Y-intercept
The y-intercept is a crucial feature of a line. It tells us where the line crosses the \(y\)-axis. This is the point at which \(x = 0\). In the equation \(y = mx + b\), \(b\) stands for the y-intercept. To find it, you plug in the slope \(m\) and the coordinates of a point on the line into the equation and solve for \(b\). Let's apply this to our calculated slope and one of the points:
  • Substitute \(m = 0\) and the point \((1, 4)\) into \(y = mx + b\).
  • The setup is: \(4 = 0 \times 1 + b\).
  • Solving gives \(b = 4\).
This \(b\) value of 4 tells us that the line crosses the \(y\)-axis at the point \((0, 4)\). This means no matter how much \(x\) changes, \(y\) remains 4.
Equation of a Line
Finally, we're ready to write the equation of the line. When given the slope \(m\) and the y-intercept \(b\), it's easy to plug these into the format \(y = mx + b\). This equation shows the relationship between any \(x\) and \(y\) on the line.
  • In our example, with \(m = 0\) and \(b = 4\), the line equation becomes: \(y = 0x + 4\).
  • Since \(0x\) equates to zero, the simplified equation is \(y = 4\).
This indicates that for all values of \(x\), \(y\) is always 4. Essentially, it describes a horizontal line extended indefinitely to the left and right across the \(y = 4\) level. In mathematical terms, this is a constant function since it keeps \(y\) value unchanged throughout its domain.

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Most popular questions from this chapter

Find the area of the triangle having the given vertices. $$(-1,2),(2,2),(-2,4)$$

Use the determinant of the coefficient matrix to determine whether the system of linear equations has a unique solution $$\begin{aligned} x_{1}+x_{2}-x_{3} &=4 \\ 2 x_{1}-x_{2}+x_{3} &=6 \\ 3 x_{1}-2 x_{2}+2 x_{3} &=0 \end{aligned}$$

Find the adjoint of the matrix \(A .\) Then use the adjoint to find the inverse of \(A,\) if possible. $$A=\left[\begin{array}{rrr} 0 & 1 & 1 \\ 1 & 2 & 3 \\ -1 & -1 & -2 \end{array}\right]$$

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