Problem 19
Question
Write the point-slope form of the equation of the line satisfying each of the conditions in Exercises \(1-28 .\) Then use the point-slope form of the equation to write the slope-intercept form of the equation. Passing through \((-3,-1)\) and \((2,4)\)
Step-by-Step Solution
Verified Answer
The point-slope form of the equation of the line is \(y + 1 = x + 3\), and the slope-intercept form of the equation of the line is \(y = x + 2\).
1Step 1: Find the slope of the line
Use the formula for the slope of a line, \(m = (y2 - y1)/(x2 - x1)\), where \((x1, y1) = (-3, -1)\) and \((x2, y2) = (2, 4)\). Substituting these points in gives \(m = (4 - (-1))/(2 - (-3)) = 5/5 = 1\). So, the slope of the line is 1.
2Step 2: Formulate the point-slope form of the equation
Next, write the point-slope form of the equation of a line \(y - y1 = m(x - x1)\) substituting \(m = 1\), and the coordinates of one point, \((x1, y1) = (-3, -1)\) into the equation. Doing this gives the point-slope form \(y - (-1) = 1(x - (-3))\), which simplifies to \(y + 1 = x + 3\).
3Step 3: Transform the equation into slope-intercept form
Finally, rewrite the point-slope form to the slope-intercept form, \(y = mx + b\), by isolating \(y\). This gives \(y = x + 3 - 1\), which simplifies to \(y = x + 2\). Therefore, the slope-intercept form of the equation is \(y = x + 2\).
Key Concepts
Slope of a LineSlope-Intercept FormAlgebraic Equations
Slope of a Line
Understanding the slope of a line is critical in algebra and geometry as it describes the steepness and direction of the line. It's often represented by the letter 'm' and can be calculated using the coordinates of two points on the line. The formula to find the slope is given by \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
To calculate the slope between two points, say \( (-3,-1) \) and \( (2,4) \) as in our exercise, we subtract the y-coordinate of the first point from the y-coordinate of the second point and do the same for the x-coordinates. This gives us \( m = \frac{4 - (-1)}{2 - (-3)} = \frac{5}{5} = 1 \), indicating a line that tilts upwards to the right at a 45-degree angle. A positive slope such as this signifies an increase in 'y' as 'x' increases.
In contrast, a negative slope indicates a decrease in 'y' with an increase in 'x', resulting in a line tilting downwards to the right. If the slope is zero, the line is horizontal, and an undefined slope corresponds to a vertical line.
To calculate the slope between two points, say \( (-3,-1) \) and \( (2,4) \) as in our exercise, we subtract the y-coordinate of the first point from the y-coordinate of the second point and do the same for the x-coordinates. This gives us \( m = \frac{4 - (-1)}{2 - (-3)} = \frac{5}{5} = 1 \), indicating a line that tilts upwards to the right at a 45-degree angle. A positive slope such as this signifies an increase in 'y' as 'x' increases.
In contrast, a negative slope indicates a decrease in 'y' with an increase in 'x', resulting in a line tilting downwards to the right. If the slope is zero, the line is horizontal, and an undefined slope corresponds to a vertical line.
Slope-Intercept Form
The slope-intercept form of a line is perhaps the most familiar algebraic representation of a line. Written as \( y = mx + b \), where 'm' is the slope and 'b' is the y-intercept - the point where the line crosses the y-axis. This form is particularly useful as it allows you to quickly sketch the behavior of the line on a graph by identifying its steepness and y-intercept directly.
Once you have the slope of a line from two points, as we calculated earlier to be 1 (or any given slope), and a point on the line, you can easily write the equation in slope-intercept form. In our exercise, we rearranged the point-slope format \( y + 1 = x + 3 \) to achieve the slope-intercept form \( y = x + 2 \), indicating that the line crosses the y-axis at (0,2).
Additionally, considering straight lines in different contexts, the slope-intercept form quickly allows one to determine how 'y' changes with 'x', which can symbolize diverse relationships, like speed versus time or cost versus quantity in real-world scenarios.
Once you have the slope of a line from two points, as we calculated earlier to be 1 (or any given slope), and a point on the line, you can easily write the equation in slope-intercept form. In our exercise, we rearranged the point-slope format \( y + 1 = x + 3 \) to achieve the slope-intercept form \( y = x + 2 \), indicating that the line crosses the y-axis at (0,2).
Additionally, considering straight lines in different contexts, the slope-intercept form quickly allows one to determine how 'y' changes with 'x', which can symbolize diverse relationships, like speed versus time or cost versus quantity in real-world scenarios.
Algebraic Equations
An algebraic equation is a mathematical statement that uses numbers, variables, and operation symbols to express a relationship. In the case of linear equations, which include our focus on the slope of a line and the slope-intercept form, they are fundamental for solving problems that involve finding the equation of a line given certain information.
Typically, algebraic equations require a process of simplification and manipulation to solve for a variable. In the step-by-step solution of our exercise, this involved isolating the variable 'y' after using the point-slope form. The manipulation to solve these equations hinges on understanding the properties of equality and legal algebraic operations that ensure the equation's balance is maintained.
Whether it be simple linear equations or more complex polynomial equations, mastering algebraic equations is a necessary skill for advancing in mathematics. They form the basis for describing and solving a vast array of mathematical problems and are often the starting point for more complex analysis in fields like engineering, physics, economics, and beyond.
Typically, algebraic equations require a process of simplification and manipulation to solve for a variable. In the step-by-step solution of our exercise, this involved isolating the variable 'y' after using the point-slope form. The manipulation to solve these equations hinges on understanding the properties of equality and legal algebraic operations that ensure the equation's balance is maintained.
Whether it be simple linear equations or more complex polynomial equations, mastering algebraic equations is a necessary skill for advancing in mathematics. They form the basis for describing and solving a vast array of mathematical problems and are often the starting point for more complex analysis in fields like engineering, physics, economics, and beyond.
Other exercises in this chapter
Problem 18
Find the \(x\) -intercept and the \(y\) -intercept of the graph of each equation. Do not graph the equation. $$2 x=4 y-13$$
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In Exercises \(13-26,\) begin by solving the linear equation for \(y .\) This will put the equation in slope-intercept form. Then find the slope and the \(y\) -
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Use intercepts and a checkpoint to graph each equation. $$x+y=5$$
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