Definition
The point-slope form of a linear equation is expressed as:
y - y1 = m(x - x1)
Here:
- (x1, y1) denotes a specific point on the line.
- m symbolizes the slope of the line.
This form is particularly beneficial when you are given the slope of a line and a point through which the line passes. It provides a straightforward method to write the equation of a line, which can then be transformed into other forms if needed, such as the slope-intercept form.
Deriving the Point-Slope Form
The point-slope form can be derived from the concept of slope itself. Recall that the slope (m) is defined as the ratio of the vertical change (Δy) to the horizontal change (Δx) between any two points on a line:
m = Δy/Δx
If we take a specific point on the line, (x1, y1), and any other point (x, y), we can express the slope as:
m = (y - y1)/(x - x1)
Rearranging this equation gives us the point-slope form:
y - y1 = m(x - x1)
Applications
Writing Equations of Lines
One of the primary applications of the point-slope form is to write the equation of a line when given a point and the slope.
Example 1:
Given the point A(3, 4) and slope m = 2, find the equation of the line.
Using the point-slope form:
y - 4 = 2(x - 3)
y - 4 = 2x - 6
y = 2x - 2
Formulating Equations from Graphs
The point-slope form is also instrumental in formulating equations directly from graphs, where a point and the slope can be identified visually.
Example 2:
If a line passes through the point B(2, -3) and has a slope of -4, determine the equation of the line.
Using the point-slope form:
y + 3 = -4(x - 2)
y + 3 = -4x + 8
y = -4x + 5
Parallel and Perpendicular Lines
In the context of parallel and perpendicular lines, the point-slope form can be employed to swiftly derive equations.
- Parallel Lines: Two lines are parallel if they have the same slope but different y-intercepts. If a line is parallel to another line with a known slope, and it passes through a given point, the point-slope form can be used to find its equation.
- Perpendicular Lines: Two lines are perpendicular if the product of their slopes is -1. If a line is perpendicular to another line with a known slope m, it will have a slope of -1/m. Using this new slope and a given point, the point-slope form can be used to find its equation.
Example 3:
Find the equation of the line perpendicular to the line y = 3x + 1 and passing through the point C(-1, 2).
The slope of the given line is 3. Therefore, the slope of the line perpendicular to it will be -1/3. Using the point-slope form:
y - 2 = (-1/3)(x + 1)
y - 2 = (-1/3)x - 1/3
y = (-1/3)x + 5/3
Analyzing Linear Models
In real-world scenarios, the point-slope form can be applied to analyze linear models by formulating equations that describe various phenomena, such as economic trends, biological growth, or physical movement, given a particular point and rate of change (slope).
Example 4:
If a company’s profit increases by £5000 for each additional product sold (slope) and the profit was £20,000 when 3 products were sold (point), find the linear model describing the profit.
Using the point-slope form:
P - 20000 = 5000(Pn - 3)
P - 20000 = 5000Pn - 15000
P = 5000Pn + 5000
Where P represents the profit and Pn represents the number of products sold.
FAQ
While the point-slope form is quite versatile in representing linear relationships, it does have limitations, particularly in its applicability to only linear models. It cannot represent non-linear, exponential, or logarithmic relationships, limiting its use in scenarios where the rate of change is not constant. Moreover, it may not provide a comprehensive view in multivariable contexts or where relationships between variables are more complex and intertwined. In such cases, advanced modelling techniques and forms might be required to accurately represent and analyse the relationships.
The point-slope form is particularly advantageous for transitioning to other forms of linear equations due to its straightforward structure. For instance, to convert the point-slope form, y - y1 = m(x - x1), to the slope-intercept form, y = mx + c, one can simply solve for y to find the y-intercept (c). This transition allows for easy graphing and further analysis of the linear relationship. Similarly, it can be rearranged to derive the standard form of the linear equation, providing various perspectives and approaches to analyse and utilise the linear relationship in mathematical and applied contexts.
The point-slope form, by definition, is specifically designed for linear relationships, where the rate of change (slope) is constant across all points on the line. Non-linear relationships, which involve curves and varying rates of change, cannot be accurately represented using the point-slope form. For non-linear relationships, other forms and methods, such as polynomial, exponential, or logarithmic models, are more appropriate to describe and analyse the varying relationship between variables.
In economics, the point-slope form can be instrumental in analysing and predicting economic trends. For instance, consider a scenario where a company observes that for every additional unit of a product produced (increase in x), the profit (y) increases by a certain amount, establishing a slope (m). If a particular profit is associated with the production of a certain number of units, a point on the graph (x1, y1) is identified. The point-slope form, y - y1 = m(x - x1), allows economists to formulate a linear model that can predict profit for different levels of production, aiding in strategic planning and decision-making.
The point-slope form provides a linear equation that essentially is a linear function, mapping each input (or x-value) to a unique output (or y-value). This linear function showcases a constant rate of change, represented by the slope (m), and can be utilised to predict y-values for any given x-value within the domain of the function. Thus, the point-slope form not only gives an equation of a line but also defines a linear function that can be used for various analytical and predictive purposes in mathematics and related fields.
Practice Questions
The point-slope form of the equation of a line is given by y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Substituting the given point D(4, 7) and slope m = -3 into the equation, we get y - 7 = -3(x - 4). Simplifying, we have y - 7 = -3x + 12. To convert this into slope-intercept form (y = mx + c), we rearrange to get y = -3x + 19. Therefore, the equation of the line in slope-intercept form is y = -3x + 19.
To find the slope (m) of the line passing through points E(-2, 5) and F(3, -4), we use the formula m = (y2 - y1) / (x2 - x1). Substituting the coordinates of points E and F, we get m = (-4 - 5) / (3 + 2) = -9/5. Now, using point E(-2, 5) in the point-slope form equation y - y1 = m(x - x1), we get y - 5 = -9/5(x + 2). Simplifying, we have y - 5 = (-9/5)x - 18/5. To convert this into slope-intercept form, we rearrange to get y = (-9/5)x + 7/5. Therefore, the equation of the line in slope-intercept form is y = (-9/5)x + 7/5.
