Problem 36
Question
Use the given conditions to write an equation for each line in point-slope form and slope-intercept form. Passing through (1,-3) with \(x\) -intercept \(=-1\)
Step-by-Step Solution
Verified Answer
The point-slope form of the line is \(y + 3 = -1.5x + 1.5\) and the slope-intercept form of the line is \(y = -1.5x - 1.5\).
1Step 1: Find the slope
The slope of a line can be found using the given x-intercept and y-coordinate of the given point. Since the line intercepts the x-axis at (-1,0), the slope \(m\) can be found using the formula \[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - (-3)}{-1 - 1} = \frac{3}{-2} = -1.5\]
2Step 2: Write the point-slope form of the equation
The point-slope form of the line is \[y - y_1 = m(x - x_1)\]. Substituting \(m = -1.5\), \(x_1 = 1\) and \(y_1 = -3\), we obtain: \[y - (-3) = -1.5(x - 1)\] which simplifies to \[y + 3 = -1.5x + 1.5\].
3Step 3: Write the slope-intercept form of the equation
The slope-intercept form of the line is \(y = mx + b\). Using the point-slope equation from step 2, rearrange to the form \(y = mx + b\): \[y = -1.5x + 1.5 - 3\] which simplifies to \[y = -1.5x - 1.5\].
Key Concepts
Slope-Intercept FormLinear EquationsSlope of a Line
Slope-Intercept Form
The slope-intercept form is an essential part of algebra, particularly when we deal with linear equations. It's a direct way to express a straight line on a coordinate plane and is written as:
\[ y = mx + b \]
Here, \( m \) represents the slope of the line, which indicates how steep the line is, and \( b \) is the y-intercept, where the line crosses the y-axis.
This form is extremely useful because it gives you almost everything you need to know about a line — its steepness and where it intersects the y-axis. By simply knowing the slope and the y-intercept, we can effortlessly sketch the line on a graph without the need for a table of values or additional points.
For example, if you are provided with a slope of -1.5 and a y-intercept of -1.5, the equation of the line in slope-intercept form is \[ y = -1.5x -1.5 \]. This means the line goes downwards (negative slope) as it moves from left to right at a tilt of 1.5 units down for every 1 unit across, and it crosses the y-axis below the origin at -1.5.
\[ y = mx + b \]
Here, \( m \) represents the slope of the line, which indicates how steep the line is, and \( b \) is the y-intercept, where the line crosses the y-axis.
This form is extremely useful because it gives you almost everything you need to know about a line — its steepness and where it intersects the y-axis. By simply knowing the slope and the y-intercept, we can effortlessly sketch the line on a graph without the need for a table of values or additional points.
For example, if you are provided with a slope of -1.5 and a y-intercept of -1.5, the equation of the line in slope-intercept form is \[ y = -1.5x -1.5 \]. This means the line goes downwards (negative slope) as it moves from left to right at a tilt of 1.5 units down for every 1 unit across, and it crosses the y-axis below the origin at -1.5.
Linear Equations
Linear equations form the foundation for understanding how quantities relate to each other in a direct, proportional manner. These equations describe lines in the coordinate plane and can be represented in various forms, including point-slope form and slope-intercept form.
A linear equation in two variables, \( x \) and \( y \), usually looks like \[ ax + by = c \], with \( a \), \( b \), and \( c \) being constants. The graph of every linear equation will be a straight line, and each point on that line is a solution to the equation.
The beauty of linear equations is in their simplicity and the insight they provide to real-world problems, such as predicting profits or losses, understanding speed and distance relationships, or analyzing trends over time. Especially in algebra, they're the building blocks to more complex functions and an integral part of any maths curriculum.
A linear equation in two variables, \( x \) and \( y \), usually looks like \[ ax + by = c \], with \( a \), \( b \), and \( c \) being constants. The graph of every linear equation will be a straight line, and each point on that line is a solution to the equation.
The beauty of linear equations is in their simplicity and the insight they provide to real-world problems, such as predicting profits or losses, understanding speed and distance relationships, or analyzing trends over time. Especially in algebra, they're the building blocks to more complex functions and an integral part of any maths curriculum.
Slope of a Line
The slope is a measure of the 'steepness' or the incline of a line and is a crucial concept in understanding how lines behave on a graph. It is denoted by the letter \( m \) and is calculated as the change in the y-coordinate divided by the change in the x-coordinate between two distinct points on the line. In simpler terms, it's the 'rise over the run'.
The formula to find the slope given two points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
For instance, if you're working with points (1, -3) and the x-intercept (-1, 0), the slope \( m \) would be calculated as \[ m = \frac{0 - (-3)}{-1 - 1} = -1.5 \]. This negative value tells us that the line falls as it moves to the right. The slope is a constant value for a given straight line, making it a powerful tool to compare different lines regarding their angles and parallelism.
The formula to find the slope given two points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
For instance, if you're working with points (1, -3) and the x-intercept (-1, 0), the slope \( m \) would be calculated as \[ m = \frac{0 - (-3)}{-1 - 1} = -1.5 \]. This negative value tells us that the line falls as it moves to the right. The slope is a constant value for a given straight line, making it a powerful tool to compare different lines regarding their angles and parallelism.
Other exercises in this chapter
Problem 36
Find \(f+g, f-g,\) fg, and \(\frac{f}{x}\). Determine the domain for each function. $$f(x)=6 x^{2}-x-1, g(x)=x-1$$
View solution Problem 36
Evaluate each function at the given values of the independent variable and simplify. \(f(x)=\frac{4 x^{3}+1}{x^{3}}\) a. \(f(2)\) b. \(f(-2)\) c. \(f(-x)\)
View solution Problem 37
Write the standard form of the equation of the circle with the given center and radius. Center \((-3,-1), r=\sqrt{3}\)
View solution Problem 37
Let \(P(x, y)\) be a point on the graph of \(y=\sqrt{x}\). Express the distance, \(d,\) from \(P\) to (1,0) as a function of the point's \(x\) -coordinate.
View solution