Problem 35
Question
Exer. 33-36: Find the slope-intercept form of the line that satisfies the given conditions. $$ \text { Through } A(5,2) \text { and } B(-1,4) $$
Step-by-Step Solution
Verified Answer
The slope-intercept form is \(y = -\frac{1}{3}x + \frac{11}{3}\).
1Step 1: Identify the Slope Formula
To find the slope of the line passing through points \(A(5,2)\) and \(B(-1,4)\), we use the slope formula: \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Here, \((x_1, y_1) = (5, 2)\) and \((x_2, y_2) = (-1, 4)\).
2Step 2: Calculate the Slope
Substitute the coordinates into the slope formula: \[m = \frac{4 - 2}{-1 - 5} = \frac{2}{-6} = -\frac{1}{3}\]. So, the slope \(m\) is \(-\frac{1}{3}\).
3Step 3: Use Point-Slope Form
To find the equation of the line, use the point-slope form \(y - y_1 = m(x - x_1)\). Choose point \(A(5,2)\) and substitute \(m = -\frac{1}{3}\), \(x_1 = 5\), and \(y_1 = 2\): \[y - 2 = -\frac{1}{3}(x - 5)\].
4Step 4: Simplify to Slope-Intercept Form
Distribute the slope on the right: \[y - 2 = -\frac{1}{3}x + \frac{5}{3}\]. Add 2 to both sides to isolate \(y\): \[y = -\frac{1}{3}x + \frac{5}{3} + 2\].
5Step 5: Simplify Further
Convert 2 into a fraction with the same denominator \(\frac{6}{3}\) and add: \[y = -\frac{1}{3}x + \frac{5}{3} + \frac{6}{3}\]. Combine the fractions: \[y = -\frac{1}{3}x + \frac{11}{3}\]. This is the slope-intercept form.
Key Concepts
Slope CalculationPoint-Slope FormLinear Equation
Slope Calculation
Understanding how to calculate the slope is fundamental in algebra, especially in linear equations. The slope of a line is essentially a measure of its steepness or inclination. In mathematical terms, it's defined as the change in the vertical direction (often referred to as the "rise") divided by the change in the horizontal direction (known as the "run").
To find the slope between two points on a graph, we use the slope formula:
Plug the values into the formula:
This shows that the line declines gently as it moves from left to right.
To find the slope between two points on a graph, we use the slope formula:
- The formula is: \(m = \frac{y_2 - y_1}{x_2 - x_1}\)
- Here, \((x_1, y_1)\) are the coordinates of the first point and \((x_2, y_2)\) are the coordinates of the second point.
Plug the values into the formula:
- \(y_2 = 4\), \(y_1 = 2\)
- \(x_2 = -1\), \(x_1 = 5\)
This shows that the line declines gently as it moves from left to right.
Point-Slope Form
Once we have the slope, we can use the point-slope form of a line to create its equation. This form is especially handy when you know a point on the line and the slope. The point-slope form equation is given by:
\[ y - 2 = -\frac{1}{3}(x - 5) \]
This equation helps in moving directly towards the slope-intercept form, providing a straight path to understanding the line's behavior through just one known point and the slope.
- \( y - y_1 = m(x - x_1) \)
- Here, \((x_1, y_1)\) is any known point on the line, and \(m\) is the slope.
\[ y - 2 = -\frac{1}{3}(x - 5) \]
This equation helps in moving directly towards the slope-intercept form, providing a straight path to understanding the line's behavior through just one known point and the slope.
Linear Equation
A linear equation in slope-intercept form is a straightforward representation of a line on a graph. It's expressed as \(y = mx + b\), where:
By distributing \(-\frac{1}{3}\) in \(y - 2 = -\frac{1}{3}(x - 5)\), you get:
\[y - 2 = -\frac{1}{3}x + \frac{5}{3}\]
Adding 2 on both sides, and converting 2 into \(\frac{6}{3}\), gives you:
\[y = -\frac{1}{3}x + \frac{11}{3}\]
This result offers a clear picture of the line, showing how it ascends or descends as the x-values change, and where it interacts with the vertical axis.
- \(m\) is the slope of the line, indicating its direction and steepness.
- \(b\) is the y-intercept, the point where the line crosses the y-axis.
By distributing \(-\frac{1}{3}\) in \(y - 2 = -\frac{1}{3}(x - 5)\), you get:
\[y - 2 = -\frac{1}{3}x + \frac{5}{3}\]
Adding 2 on both sides, and converting 2 into \(\frac{6}{3}\), gives you:
\[y = -\frac{1}{3}x + \frac{11}{3}\]
This result offers a clear picture of the line, showing how it ascends or descends as the x-values change, and where it interacts with the vertical axis.
Other exercises in this chapter
Problem 35
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