3.6 Writing the Equation of a Line Given Two Points

Scott McDaniel

3.6 Writing the Equation of a Line Given Two Points

Learning Objectives

After completing this section, you should be able to:

Find the slope of a line given two points.
Write the equation of a line in slope-intercept form ( $y = mx + b$ ).
Use slope-intercept form to find equations of parallel and perpendicular lines.
Recognize and write equations of vertical and horizontal lines.

Two Methods for Writing a Linear Equation

There are two common ways to find the equation of a line:

Slope-Intercept Form: Substitute the slope (m) and a known point ((x, y)) into $y = mx + b$ , then solve for b.
Point-Slope Form: Use $y - y_1 = m(x - x_1)$ , where m is the slope and (x_1, y_1) is a point on the line.

Both methods lead to the same final equation.

Example 3.6.1 — Two methods

Find the equation of the line with slope 3 through $(4,1)$ .

Method 1: Using slope-intercept form

When you know one point and the slope, start with $y = mx + b$ . Substitute the slope for $m$ and the point $(x,y)$ to find $b$ .

We know $x=4$ , $y=1$ , and $m=3$ . Substitute into $y = mx + b$ and solve for $b$ .

Solution (Method 1)

$y = mx + b$

$1 = 3(4) + b$

$1 = 12 + b$

$b = -11$

Equation: $y = 3x - 11$

Method 2: Using point-slope form

Substitute $(x_1,y_1)=(4,1)$ and $m=3$ into $y-y_1 = m(x-x_1)$ .

Solution (Method 2)

$y - 1 = 3(x - 4)$

$y - 1 = 3x - 12$

$y = 3x - 11$

This matches the equation from Method 1.

Optional: Where point-slope form comes from

Start with the slope formula using points $(x_1,y_1)$ and $(x,y)$ on the line:

$m=\frac{y-y_1}{x-x_1}$

Multiply both sides by $(x-x_1)$ :

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

Rewriting gives point-slope form:

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

Example 3.6.2

Find the equation of the line with slope $-2$ through $(1,4)$ .

Solution

Start with $y = mx + b$ :

$y = -2x + b$

Substitute $(1,4)$ :

$4 = -2(1) + b$

$4 = -2 + b$ → $b=6$

Equation: $y = -2x + 6$

Try It

Finding the Equation Given Two Points

When two points are given, first find the slope:

$m=\frac{y_2-y_1}{x_2-x_1}$

Then use one point and the slope to solve for $b$ in $y = mx + b$ .

Example 3.6.3

Find the equation of the line through $(2,4)$ and $(3,7)$ .

Solution

Step 1: Find the slope

$m=\frac{7-4}{3-2}=\frac{3}{1}=3$

Step 2: Solve for b using one point

$y=mx+b$

$4=3(2)+b$ → $4=6+b$ → $b=-2$

Equation: $y=3x-2$

Example 3.6.4

Find the equation of the line through $(-3,5)$ and $(2,8)$ .

Solution

Step 1: Find the slope

$m=\frac{8-5}{2-(-3)}=\frac{3}{5}$

Step 2: Solve for b using (2, 8)

$8=\frac{3}{5}(2)+b$ → $8=\frac{6}{5}+b$

$b=8-\frac{6}{5}=\frac{40}{5}-\frac{6}{5}=\frac{34}{5}$

Equation: $y=\frac{3}{5}x+\frac{34}{5}$

Try It

Writing Equations of Parallel and Perpendicular Lines

Parallel lines have the same slope.

Perpendicular lines have slopes that are negative reciprocals.

Example 3.6.5 — Parallel line

Write the equation of a line parallel to $y=4x-7$ and passing through $(-3,1)$ .

Solution

Parallel lines have the same slope, so $m=4$ .

Graph of line y = 4x-7 and the point (-3,1) — **Figure 3.6E**

Use $y=mx+b$ with point $(-3,1)$ :

$1=4(-3)+b$ → $1=-12+b$ → $b=13$

Equation: $y=4x+13$

Graph of both lines: y = 4x-7 and y = 4x + 13 (showing the line going through the point (-3,1) — **Figure 3.6F**

Example 3.6.6 — Perpendicular line

Write the equation of a line perpendicular to $y=4x-7$ and passing through $(-3,1)$ .

Solution

The perpendicular slope is the negative reciprocal of 4, so $m=-\frac{1}{4}$ .

Use $y=mx+b$ with point $(-3,1)$ :

$1=\left(-\frac{1}{4}\right)(-3)+b$ → $1=\frac{3}{4}+b$

$b=1-\frac{3}{4}=\frac{1}{4}$

Equation: $y=-\frac{1}{4}x+\frac{1}{4}$

Graph of the original line y = 4x - 7 and the graph of the line perpendicular y = (-1/4)x + 1/4 — **Figure 3.6H**

Try It

Special Cases: Vertical and Horizontal Lines

**Table 3.6**
Line Type	Description	Example
Horizontal	$y = \text{constant}$	$y=2$
Vertical	$x = \text{constant}$	$x=-3$

Quick Reminder

Horizontal lines have slope $0$ . Vertical lines have undefined slope.

Example 3.6.7 — Horizontal and vertical lines

a) Find the equation of the line through $(0,2)$ and $(4,2)$ .

Solution

Both points have the same y-value, so the line is horizontal.

Equation: $y=2$

b) Find the equation of the line through $(-1,-3)$ and $(-1,4)$ .

Solution

Both points have the same x-value, so the line is vertical.

Equation: $x=-1$

Note: You don’t need to plot points for these equations. Just remember that $y=\text{number}$ is a horizontal line and $x=\text{number}$ is a vertical line.

Quick Tip: Horizontal lines: $y=\text{number}$ Vertical lines: $x=\text{number}$

Key Takeaways

To find slope: $m=\frac{y_2-y_1}{x_2-x_1}$
To find $b$ : substitute one point into $y=mx+b$
Parallel lines → same slope
Perpendicular lines → negative reciprocal slopes
Horizontal lines → $y=\text{constant}$ (slope 0)
Vertical lines → $x=\text{constant}$ (undefined slope)

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Two Methods for Writing a Linear Equation

Finding the Equation Given Two Points

Writing Equations of Parallel and Perpendicular Lines

Special Cases: Vertical and Horizontal Lines

Quick Reminder

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