**Introduction to The Equation of a Line**

The equation of a line in the plane will describe—in compact form—all the points that lie on that line. We determine the equation of a given line by writing its slope in two different ways and then equating them. Some examples best illustrate the idea.

**Example 1**

Determine the equation of the line with slope 3 that passes through the point (2, 1).

**Solution 1**

Let ( *x* , *y* ) be a variable point on the line. Then we can use that variable point together with (2, 1) to calculate the slope:

On the other hand, we are given that the slope is *m* = 3. We may equate the two expressions for slope to obtain

This may be simplified to *y* = 3 *x* − 5.

**Math Note:** The form *y* = 3 *x* − 5 for the equation of a line is called the *slope-intercept form* . The slope is 3 and the line passes through (0, 5) (its *y* -intercept).

**Math Note:** Equation (*) may be rewritten as *y* − 1 = 3( *x* − 2). In general, the line with slope *m* that passes through the point ( *x* _{0} , *y* _{0} ) can be written as *y* − *y* _{0} = *m* ( *x* − *x* _{0} ). This is called the *point-slope* form of the equation of a line.

**You Try It:** Write the equation of the line that passes through the point (−3, 2) and has slope 4.

**Example 2**

Write the equation of the line passing through the points (−4, 5) and (6, 2).

**Solution 2**

Let ( *x* , *y* ) be a variable point on the line. Using the points ( *x* , *y* ) and (−4, 5), we may calculate the slope to be

On the other hand, we may use the points (−4, 5) and (6, 2) to calculate the slope:

Equating the two expressions for slope, we find that

Simplifying this identity, we find that the equation of our line is

**You Try It:** Find the equation of the line that passes through the points (2, −5) and (−6, 1).

In general, the line that passes through points ( *x* _{0} , *y* _{0} ) and ( *x* _{1} , *y* _{1} ) has equation

This is called the *two-point form* of the equation of a line.

**Example 3**

Find the line perpendicular to *y* = 3 *x* − 6 that passes through the point (5, 4).

**Solution 3**

We know from the Math Note immediately after Example 1.10 that the given line has slope 3. Thus the line we seek (the perpendicular line) has slope −1/3. Using the point-slope form of a line, we may immediately write the equation of the line with slope −1/3 and passing through (5, 4) as

**Summary:**

We determine the equation of a line in the plane by finding two expressions for the slope and equating them.

If a line has slope *m* and passes through the point ( *x* _{0} , *y* _{0} ) then it has equation

*y* − *y* _{0} = *m* ( *x* − *x* _{0} ).

This is the point-slope form of a line.

If a line passes through the points ( *x* _{0} , *y* _{0} ) and ( *x* _{1} , *y* _{1} ) then it has equation

This is the two-point form of a line.

**You Try It:** Find the line perpendicular to 2 *x* + 5 *y* = 10 that passes through the point (1, 1). Now find the line that is parallel to the given line and passes through (1, 1).

Practice problems for this concept can be found at: Calculus Basics Practice Test.

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