Lesson 3.3: Find the Equation of a Line

New Concept

This lesson covers how to determine a line's equation from various starting points: using slope and a point, two points, or its relationship (parallel/perpendicular) to another line. This skill is crucial for modeling real-world linear relationships.

What’s next

You've got the big picture. Now, you'll work through interactive examples and practice cards to master creating linear equations for any scenario.

Property

To find the equation of a line when given the slope and $y$-intercept, use the slope-intercept form of a linear equation, $y = mx + b$. Substitute the given slope for $m$ and the $y$-coordinate of the $y$-intercept for $b$.

Examples

• Find the equation of a line with slope $m = 5$ and $y$-intercept $(0, 2)$. We substitute the values into $y = mx + b$ to get $y = 5x + 2$.
• Find the equation of a line with slope $m = -\frac{1}{2}$ and $y$-intercept $(0, -3)$. We substitute the values into $y = mx + b$ to get $y = -\frac{1}{2}x - 3$.
• Find the equation of a line with slope $m = 0$ and $y$-intercept $(0, 7)$. We substitute the values into $y = mx + b$ to get $y = 0x + 7$, which simplifies to $y = 7$.

Explanation

This is the most direct way to write a line's equation. If you know its steepness (slope) and where it crosses the vertical axis (y-intercept), you can plug those two values directly into the slope-intercept formula.

Property

The point-slope form of an equation of a line with slope $m$ and containing the point $(x_1, y_1)$ is:

Examples

• For a line with slope $m=4$ passing through $(2, 5)$, the point-slope form is $y - 5 = 4(x - 2)$.
• For a line with slope $m=-3$ passing through $(-1, 6)$, the point-slope form is $y - 6 = -3(x - (-1))$, which simplifies to $y - 6 = -3(x + 1)$.
• For a line with slope $m=\frac{2}{5}$ passing through $(10, -3)$, the point-slope form is $y - (-3) = \frac{2}{5}(x - 10)$, which simplifies to $y + 3 = \frac{2}{5}(x - 10)$.

Explanation

Think of the point-slope form as a recipe for a line's equation. It's incredibly useful when you don't know the y-intercept. All you need is the slope and any single point that the line passes through.

Property

To find an equation of a line given a point $(x_1, y_1)$ and the slope $m$:

1. Identify the slope $m$ and the point $(x_1, y_1)$.
2. Substitute the values into the point-slope form, $y - y_1 = m(x - x_1)$.
3. Write the equation in slope-intercept form, $y = mx + b$, by solving for $y$.

Examples

• Find the equation of a line with slope $m=2$ that contains $(3, 4)$. Using point-slope form: $y - 4 = 2(x - 3)$. Distributing gives $y - 4 = 2x - 6$. The final equation is $y = 2x - 2$.
• Find the equation of a line with slope $m = -\frac{1}{4}$ that contains $(8, -1)$. Using point-slope form: $y - (-1) = -\frac{1}{4}(x - 8)$. This becomes $y + 1 = -\frac{1}{4}x + 2$. The final equation is $y = -\frac{1}{4}x + 1$.
• Find the equation of a horizontal line containing $(-4, 9)$. A horizontal line has slope $m=0$. Using point-slope form: $y - 9 = 0(x - (-4))$. This simplifies to $y - 9 = 0$, so the equation is $y = 9$.

Explanation

This method uses the point-slope form as a starting point. After plugging in the slope and the point's coordinates, you simplify and rearrange the equation to get it into the familiar $y = mx + b$ format.

Property

To find the equation of a line given two points $(x_1, y_1)$ and $(x_2, y_2)$:

1. Find the slope using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.
2. Choose one of the given points.
3. Substitute the slope and the chosen point into the point-slope form, $y - y_1 = m(x - x_1)$.
4. Write the equation in slope-intercept form.

Examples

• Find the equation of a line containing $(1, 2)$ and $(3, 8)$. First, find the slope: $m = \frac{8 - 2}{3 - 1} = \frac{6}{2} = 3$. Using point $(1, 2)$: $y - 2 = 3(x - 1)$, which simplifies to $y = 3x - 1$.
• Find the equation of a line containing $(-2, 5)$ and $(4, 2)$. The slope is $m = \frac{2 - 5}{4 - (-2)} = \frac{-3}{6} = -\frac{1}{2}$. Using point $(4, 2)$: $y - 2 = -\frac{1}{2}(x - 4)$, which simplifies to $y = -\frac{1}{2}x + 4$.
• Find the equation of a line containing $(5, 1)$ and $(5, -4)$. The slope is $m = \frac{-4 - 1}{5 - 5} = \frac{-5}{0}$, which is undefined. This is a vertical line. The equation is $x=5$.

Explanation

If you have two points, you can first determine the line's slope. Then, pick either of the two points and use the 'slope and a point' method to find the final equation in slope-intercept form.

Property

To find the equation of a line parallel to a given line that contains a given point:

1. Find the slope of the given line.
2. The slope of the parallel line is the same, $m_{\parallel} = m$.
3. Use the point-slope form with the given point and this slope to find the new equation.
4. Write the equation in slope-intercept form.

Examples

• Find a line parallel to $y = 3x + 1$ that contains the point $(2, 7)$. The slope is $m=3$. Using point-slope form: $y - 7 = 3(x - 2)$. This simplifies to $y = 3x + 1$.
• Find a line parallel to $4x + 2y = 8$ that contains $(1, -1)$. First, solve for y: $2y = -4x + 8$, so $y = -2x + 4$. The slope is $m=-2$. Using point-slope form: $y - (-1) = -2(x - 1)$. This simplifies to $y = -2x + 1$.
• Find a line parallel to $y = 5$ that contains $(4, -2)$. The line $y=5$ is horizontal with slope $m=0$. The parallel line is also horizontal, so its equation is $y = -2$.

Explanation

Parallel lines run in the same direction, so they have the exact same slope. Find the slope of the first line, then use that same slope with the new point to create the equation for the second line.

Property

To find the equation of a line perpendicular to a given line that contains a given point:

1. Find the slope of the given line, $m$.
2. The slope of the perpendicular line is the negative reciprocal, $m_{\perp} = -\frac{1}{m}$.
3. Use the point-slope form with the given point and the perpendicular slope.
4. Write the equation in slope-intercept form.

Examples

• Find a line perpendicular to $y = 2x - 5$ that contains $(4, 1)$. The original slope is $m=2$. The perpendicular slope is $m_{\perp} = -\frac{1}{2}$. Using point-slope form: $y - 1 = -\frac{1}{2}(x - 4)$. This simplifies to $y = -\frac{1}{2}x + 3$.
• Find a line perpendicular to $y = -\frac{1}{3}x + 2$ that contains $(1, 5)$. The original slope is $m=-\frac{1}{3}$. The perpendicular slope is $m_{\perp} = 3$. Using point-slope form: $y - 5 = 3(x - 1)$. This simplifies to $y = 3x + 2$.
• Find a line perpendicular to the vertical line $x = 4$ that contains $(1, -2)$. A line perpendicular to a vertical line is a horizontal line, which has a slope of $m=0$. The equation is simply $y = -2$.

Explanation

Perpendicular lines intersect at a right angle. Their slopes are 'negative reciprocals' – you flip the fraction and change the sign. Use this new slope with the given point to find the perpendicular line's equation.