Problem 37
Question
Find an equation of the line that passes through the point and has the indicated slope. Then sketch the line. Point \(\quad\) Slope \(\left(4, \frac{5}{2}\right) \quad m=\frac{4}{3}\)
Step-by-Step Solution
Verified Answer
The equation of the line is \(y = 4/3x - 16/3 + 5/2\). The line passes through the point (4,5/2) and has a positive slope, meaning it slants upwards from left to right when sketched.
1Step 1: Identify given point and slope
The given point is (4, 5/2) and the slope of the line (m) is 4/3.
2Step 2: Use point-slope form of line equation
The point-slope form of the equation of a line is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) are the coordinates of the given point and m is the slope. Substituting the given values, the equation becomes \(y - 5/2 = 4/3 * (x - 4)\)
3Step 3: Simplify equation
Distribute the right side to get \(y - 5/2 = 4/3*x - 16/3\). Then, add 5/2 to both sides to isolate y. The final equation is \(y = 4/3*x - 16/3 + 5/2\)
4Step 4: Plot the line
Start by plotting the point (4,5/2), then use the slope (rise over run) to find another point and draw a straight line through the points. The line should slope upward because the slope is positive.
Key Concepts
Point-Slope FormLinear EquationsSlope-Intercept Form
Point-Slope Form
Understanding how to form an equation using the point-slope form is a fundamental part of algebra and it is crucial for graphing linear equations effectively. In our exercise, we were asked to determine the equation of a line passing through a given point with a specified slope. The point-slope form is written as:
\( y - y_1 = m(x - x_1) \)
where \( (x_1, y_1) \) represent the coordinates of the point on the line, and \( m \) is the slope of the line. The slope quantifies the steepness and direction of the line. To use this form to write the equation, one must first identify the coordinates of the given point and the slope, then substitute these values into the formula.
For clarity, let's break it down using our exercise example. We were given the point \( (4, \frac{5}{2}) \) and the slope \( \frac{4}{3} \). By substituting these into the formula, we obtained:
\( y - \frac{5}{2} = \frac{4}{3}(x - 4) \)
Once we have the equation in point-slope form, it can be used to graph the line or converted into different forms, such as the slope-intercept form, for various applications.
\( y - y_1 = m(x - x_1) \)
where \( (x_1, y_1) \) represent the coordinates of the point on the line, and \( m \) is the slope of the line. The slope quantifies the steepness and direction of the line. To use this form to write the equation, one must first identify the coordinates of the given point and the slope, then substitute these values into the formula.
For clarity, let's break it down using our exercise example. We were given the point \( (4, \frac{5}{2}) \) and the slope \( \frac{4}{3} \). By substituting these into the formula, we obtained:
\( y - \frac{5}{2} = \frac{4}{3}(x - 4) \)
Once we have the equation in point-slope form, it can be used to graph the line or converted into different forms, such as the slope-intercept form, for various applications.
Linear Equations
Linear equations are the simplest form of equations to understand in algebra. They are used to represent straight lines on a Cartesian plane. A linear equation in two variables typically looks like:
\( ax + by = c \)
These equations help us build a relationship between the x and y values, signifying how the value of y changes with x. The coefficients \( a \) and \( b \) determine the slope and intercept of the line.
In our problem, after setting up the equation using the point-slope form, we can distribute and isolate to get a clearer linear equation which maps out exactly how y depends on x. The slope \( m \) gives us the 'rise over run,' and when we plot this on the graph starting from our given point, we can see the line take shape. This is the core value of understanding linear equations; they model real-world situations where relationships between variables can be approximated linearly, such as speed, distance, and time.
\( ax + by = c \)
These equations help us build a relationship between the x and y values, signifying how the value of y changes with x. The coefficients \( a \) and \( b \) determine the slope and intercept of the line.
In our problem, after setting up the equation using the point-slope form, we can distribute and isolate to get a clearer linear equation which maps out exactly how y depends on x. The slope \( m \) gives us the 'rise over run,' and when we plot this on the graph starting from our given point, we can see the line take shape. This is the core value of understanding linear equations; they model real-world situations where relationships between variables can be approximated linearly, such as speed, distance, and time.
Slope-Intercept Form
The slope-intercept form is one of the most commonly used forms of a linear equation and is incredibly helpful in visualizing graphs quickly. This form is represented as:
\( y = mx + b \)
where \( m \) is the slope and \( b \) is the y-intercept, the point where the line crosses the y-axis. Starting with our previously derived point-slope form, simplify and rearrange the terms to solve for y, converting it into the slope-intercept form.
From our exercise, we distributed the slope and isolated y to get:
\( y = \frac{4}{3}x - \frac{16}{3} + \frac{5}{2} \)
With some arithmetic simplification, we can find a common denominator and combine the constants to express the equation neatly in slope-intercept form. This form immediately shows us the slope of the line, which helps understand the direction the line moves across the graph, and the y-intercept, which gives us a starting point for graphing. Being able to manipulate and understand the slope-intercept form is essential, as it enables students to sketch and interpret lines and their behaviors on a graph effortlessly.
\( y = mx + b \)
where \( m \) is the slope and \( b \) is the y-intercept, the point where the line crosses the y-axis. Starting with our previously derived point-slope form, simplify and rearrange the terms to solve for y, converting it into the slope-intercept form.
From our exercise, we distributed the slope and isolated y to get:
\( y = \frac{4}{3}x - \frac{16}{3} + \frac{5}{2} \)
With some arithmetic simplification, we can find a common denominator and combine the constants to express the equation neatly in slope-intercept form. This form immediately shows us the slope of the line, which helps understand the direction the line moves across the graph, and the y-intercept, which gives us a starting point for graphing. Being able to manipulate and understand the slope-intercept form is essential, as it enables students to sketch and interpret lines and their behaviors on a graph effortlessly.
Other exercises in this chapter
Problem 37
Sketch the graph of the function and determine whether the function is even, odd, or neither. \(f(x)=3\)
View solution Problem 37
Evaluate the function at each specified value of the independent variable and simplify. \(f(y)=3-\sqrt{y}\) (a) \(f(4)\) (b) \(f(100)\) (c) \(f\left(4 x^{2}\rig
View solution Problem 37
Check for symmetry with respect to both axes and the origin. \(x^{4}-2 y=0\)
View solution Problem 38
Determine the domain of (a) \(f\), (b) \(g\), and (c) \(f \circ g\). \(f(x)=\sqrt[3]{x+1}, \quad g(x)=x^{3}\)
View solution