Slope Intercept Form

    • Definition of slope-intercept form
    • The Slope-Intercept form
    • Derivation of the Slope Intercept Formula
    • Straight Line Equation Using Slope-Intercept Form
    • Standard Form to Slope Intercept Form
    • Important Notes on the Slope Intercept Form
    • Solved Examples
    • Practice Problems
    • Frequently Asked Questions

     

    Definition of Slope-Intercept Form

    The slope-intercept form is one of the forms used to write the equation of a straight line. It is a preferred way of writing the equation of a straight line when the slope of the line and the `y`-intercept of the line are given. The slope indicates how steep the line is and whether it rises or falls as it moves from left to right. The `y`-intercept is the point where the line crosses the `y`-axis. This form is useful in graphing linear equations and understanding their behavior.

     

    The Slope-Intercept Form

    The slope-intercept form of a linear equation is `y = mx + b`. In this form:

    `m` represents the slope of the line, indicating its steepness and direction.

    `b` represents the `y`-intercept, the point where the line crosses the `y`-axis.

     

    Example: Find the equation of the line passing through the point `(3, 5)` with a slope of `2`.

    Solution:

    We can use the slope-intercept form of a linear equation, which is `y = mx + b`.

    Given:

    • Slope \(m = 2\)
    • Point \( (x, y) = (3, 5)\)

    We can substitute the values of \(m\) into \( y = mx + b \) and write the equation of the line as

    \( y = 2x + b \)

    To find the value of \(b\), we use the given point \( (3, 5) \) and substitute its coordinates into the equation:

    \( 5 = 2(3) + b \)

    \( 5 = 6 + b \)

    \( b = 5 - 6 \)

    \( b = -1 \)

    Now, we have found the value of \(b\), which is \(-1\). 

    Therefore, the equation of the line passing through the point `(3, 5)` with a slope of `2` is:

    \( \boxed{y = 2x - 1} \)

     

    Derivation of the Slope Intercept Formula

    Let's derive the slope intercept formula for a straight line.

    Consider a line with a slope \( m \) that intersects the `y`-axis at the point \( (0, b) \), where \( b \) represents the `y`-intercept. Also, let's take an arbitrary point \( (x, y) \) on this line.

    Use the slope formula, which states that the slope \( m \) of a line joining two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:

    \( m = \frac{{y_2 - y_1}}{{x_2 - x_1}} \)

    Let's assign \( (x_1, y_1) = (0, b) \) and \( (x_2, y_2) = (x, y) \). Substituting these values into the slope formula, we get:

    \( m = \frac{{y - b}}{{x - 0}} \)

    \( \Rightarrow m = \frac{{y - b}}{{x}} \)

    Multiplying both sides by \( x \), we have:

    \( mx = y - b \)

    Adding \( b \) to both sides, we obtain:

    \( y = mx + b \)

     

    Straight Line Equation Using Slope-Intercept Form

    The equation `y = mx + b` represents the equation of a straight line involving its slope \( m \) and its `y`-intercept \( b \). Hence, this form of the equation is termed the slope-intercept form, which we derived from the given conditions.

    To determine the equation of a line using the slope-intercept form, two crucial pieces of information are needed: 

    • The inclination of the line (or its slope, represented by \( m \), or the angle it makes with the `x`-axis, denoted by \( \theta \))
    • The line's placement (specified by its `y`-intercept, \( b \), or the point on the `y`-axis through which it passes). 

    These parameters uniquely define any line.

    The steps to find the equation of a line using the slope-intercept form are as follows:

    Step `1`: Identify the `y`-intercept, denoted as \( b \), and the slope of the line, denoted as \( m \). If the slope is not directly given, you can determine it using the angle it makes with the `x`-axis, provided it is available.

    Step `2`: Apply the slope-intercept formula: \( y = mx + b \).

     

    Example: A line is inclined at an angle of \( 60^\circ \) to the horizontal axis and passes through the point \( (0, -1) \). Find the equation of this line.

    Solution: 

    Using \( m = \tan(60^\circ) = \sqrt{3} \), we can use the slope-intercept formula \( y = mx + b \).

    Thus, the equation of the line is \( y = \sqrt{3}x - 1 \).

     

    Standard Form to Slope Intercept Form

    Standard form is yet another form of writing the equation of any straight line. Converting a linear equation from standard form \( Ax + By = C \) to slope-intercept form \( y = mx + b \) involves rearranging the terms to isolate \( y \) on one side of the equation. Here are the steps to convert from standard form to slope-intercept form:

    `1`. Identify the values of \( A \), \( B \), and \( C \): These coefficients represent the parameters of the linear equation in standard form.

    `2`. Solve for \( y \): To isolate \( y \), subtract \( Ax \) from both sides of the equation:

     \( Ax + By = C \)

    \( By = -Ax + C \)

    `3`. Divide by \( B \): Divide every term by \( B \) to isolate \( y \) on one side:

    \( y = -\frac{A}{B}x + \frac{C}{B} \)

    `4`. Identify \( m \) and \( b \): The coefficient of \( x \), \( -\frac{A}{B} \), represents the slope \( m \), while \( \frac{C}{B} \) represents the `y`-intercept \( b \).

    `5`. Write the equation in slope-intercept form: Substitute the values of \( m \) and \( b \) into the equation:

    \( y = mx + b \)

    \( y = \left(-\frac{A}{B}\right)x + \frac{C}{B} \)

     

    Important Notes on the Slope Intercept Form

    `1`. Slope and `Y`-Intercept: The slope-intercept form of a linear equation \( y = mx + b \) uniquely defines a line by its slope \( m \) and `y`-intercept \( b \).

    `2`. Graphical Representation: This form simplifies graphing. To graph a linear equation given in slope intercept form, start graphing by plotting the `y`-intercept first. Next, starting from the `y`-intercept, use the slope value and follow the rise over run to plot the second point. Connect the two points using a straight line.

    `3`. Versatility: The slope-intercept form allows easy identification of key properties such as slope and intercepts, making it invaluable for understanding and analyzing linear relationships.

     

    Solved Examples

    Example 1. Find the equation of a line passing through the point \((2, 4)\) with a slope of \(3\).

    Solution:

    Given the point \((2, 4)\) and the slope \(m = 3\), we can use the slope-intercept form \(y = mx + b\).

    Substituting the given values, we get:

    \(4 = 3(2) + b\)

    \(4 = 6 + b\)

    \(b = 4 - 6\)

    \(b = -2\)

    Therefore, the equation of the line is \(y = 3x - 2\).

     

    Example `2`. Determine the equation of a line with a slope of \(-\frac{1}{2}\) passing through the point \((5, 3)\).

    Solution:

    Given the slope \(m = -\frac{1}{2}\) and the point \((5, 3)\), we use the slope-intercept form \(y = mx + b\).

    Substituting the given values:

    \(3 = -\frac{1}{2}(5) + b\)

    \(3 = -\frac{5}{2} + b\)

    \(b = 3 + \frac{5}{2}\)

    \(b = \frac{11}{2}\)

    Therefore, the equation of the line is \(y = -\frac{1}{2}x + \frac{11}{2}\).

     

    Example `3`. What is the `y`-intercept of the line with the equation `2y−4x=10x−20`?

    Solution:

    To find the `y`-intercept of the line with the equation \(2y - 4x = 10x - 20\), we need to rewrite the equation in the slope-intercept form \(y = mx + b\).

    Given equation: \(2y - 4x = 10x - 20\)

    First, let's isolate \(y\)-term to one side of the equation:

    \(2y = 4x + 10x - 20\)

    \(2y = 14x - 20\)

    Next, divide both sides by `2` to solve for \(y\):

    \(y = 7x - 10\)

    Now, we have the equation in slope-intercept form, where \(m = 7\) (the slope) and \(b = -10\) (the `y`-intercept).

    Therefore, the `y`-intercept of the line is \(-10\).

     

    Example `4`. What is the equation of the line between `(0,10)` and `(4,20)`?

    Solution:

    To find the equation of the line passing through the points \((0, 10)\) and \((4, 20)\), we first need to determine the slope (\(m\)) using the formula:

    \( m = \frac{{y_2 - y_1}}{{x_2 - x_1}} \)

    Given the points \((0, 10)\) and \((4, 20)\), we can substitute their coordinates into the formula:

    \( m = \frac{{20 - 10}}{{4 - 0}} \)

    \( m = \frac{{10}}{{4}} \)

    \( m = 2.5 \)

    Now that we have the slope (\(m\)), we can use one of the given points and the slope to find the equation of the line using the point-slope form:

    \( y - y_1 = m(x - x_1) \)

    Let's use the point \((0, 10)\):

    \( y - 10 = 2.5(x - 0) \)

    \( y - 10 = 2.5x \)

    Now, let's isolate \(y\) :

    \( y = 2.5x + 10 \)

    Therefore, the equation of the line passing through the points \((0, 10)\) and \((4, 20)\) is \( y = 2.5x + 10 \).

     

    Example 5. Rewrite the equation in slope-intercept form: `2/3y=1/2x+3`

    Solution:

    To rewrite the equation \( \frac{2}{3}y = \frac{1}{2}x + 3 \) in slope-intercept form \(y = mx + b\), we need to isolate \(y\) on one side of the equation.

    Given equation: \( \frac{2}{3}y = \frac{1}{2}x + 3 \)

    First, let's isolate \(y\) by multiplying both sides by \(\frac{3}{2}\) to get rid of the fraction:

    \( \frac{2}{3}y \times \frac{3}{2} = \frac{1}{2}x \times \frac{3}{2} + 3 \times \frac{3}{2} \)

    \( y = \frac{3}{4}x + \frac{9}{2} \)

    Now, we have the equation in slope-intercept form, where \(m = \frac{3}{4}\) (the slope) and \(b = \frac{9}{2}\) (the `y`-intercept).

    Therefore, the equation in slope-intercept form is:

    \( y = \frac{3}{4}x + \frac{9}{2} \)

     

    Practice Problems

    Q`1`. What is the equation of a line with a slope of \(2\) passing through the point \((3, 7)\)?

    1. \(y = 2x + 1\)
    2. \(y = 2x - 1\)
    3. \(y = 2x + 3\)
    4.  \(y = 2x - 3\)

    Answer: a. 

     

    Q`2`. Select the equation of a line with a slope of \(-\frac{3}{4}\) passing through the point \((6, 2)\).

    1.  \(y = -\frac{3}{4}x + 8\)
    2.  \(y = -\frac{3}{4}x + \frac{13}{2}\)
    3.  \(y = -\frac{4}{3}x + \frac{13}{2}\)
    4.  \(y = -\frac{4}{3}x + 8\)

    Answer: b. 

     

    Q`3`. Write the equation \(3x - 6y = 10\) in slope-intercept form.

    1.  \(y = \frac{1}{2}x - \frac{5}{3}\)  
    2.  \(y = \frac{1}{2}x + \frac{5}{3}\)  
    3.  \(y = -\frac{1}{2}x - \frac{5}{3}\)  
    4.  \(y = -\frac{1}{2}x + \frac{5}{3}\)

    Answer: a.

     

    Q`4`. Identify the equation given in slope-intercept form from the following choices:

    1. \(x = \frac{1}{4}y - 2\)
    2. \(y = 4x + 2\)
    3. \(4x - y + 2 = 0\)
    4. \(4x - y = -2\)

    Answer: b. 

     

    Q`5`. Rewrite the equation \(2 + 4y = 3x + 10\) in slope-intercept form.

    1.  \(y = \frac{3}{4}x + 2\)
    2. \(y = \frac{3}{4}x - \frac{1}{2}\)
    3. \(y = \frac{3}{4}x + \frac{5}{2}\)
    4. \(y = \frac{3}{4}x - \frac{5}{2}\)

    Answer: a.

     

    Frequently Asked Questions

    Q`1`. What is the slope-intercept form of a linear equation?

    Answer: The slope-intercept form of a linear equation is \(y = mx + b\), where \(m\) represents the slope of the line and \(b\) represents the `y`-intercept.

     

    Q`2`. How do you find the slope-intercept form of a line given its slope and a point?

    Answer: You can use the formula \(y = mx + b\), where \(m\) is the slope and \(b\) is the `y`-intercept. Substitute the values of the slope and the given point to find the equation of the line.

     

    Q`3`. What information does the slope-intercept form provide about a line?

    Answer: The slope-intercept form allows you to easily identify the slope of the line, which represents its steepness and direction, as well as the `y`-intercept, which indicates where the line intersects the `y`-axis.

     

    Q`4`. Can a line have a negative slope in the slope-intercept form?

    Answer: Yes, a line can have a negative slope. In the slope-intercept form, if \(m\) is negative, the line slopes downward from left to right.

     

    Q`5`. How does the slope-intercept form simplify graphing linear equations?

    Answer: The slope-intercept form directly provides the slope and `y`-intercept of the line, making it straightforward to plot the line on a graph without needing to calculate additional points.