Geometry (A): READ: Graphing Lines Reading

Equations of Lines

Learning Objectives

identify and write equations in slope-intercept form.
identify equations of parallel lines.
identify equations of perpendicular lines.
identify and write equations in standard form.

Introduction

Every line that you can represent graphically on the coordinate plane can also be represented algebraically. That means that you can create an equation relating $x$ and $y$ that corresponds to any graph of a straight line. In this lesson, you’ll learn how to create an equation from a graph or points given, identify equations of parallel and perpendicular lines, and practice using both slope-intercept and standard form.

Slope-Intercept Equations

The first type of linear equation to study is the most straightforward. It is called slope-intercept form and involves both the slope of the line and its $y-$ intercept. A $y-$ intercept is the point at which the line crosses the vertical $(y)$ axis. So, it will be the value of $y$ when $x$ is equal to $0$ . The generic formula for an equation in slope-intercept form is as follows.

$y = mx + b$

In this equation, $y$ and $x$ remain as variables, $m$ is the slope of the line, and $b$ is the $y-$ intercept of the line. So, if you know that a line has a slope of $4$ and it crosses the $y-$ axis at $(0,8),$ its equation in slope-intercept form would be $y = 4x + 8$ .

This form is especially useful for identifying the equation of a line given its graph. You already know how to deduce the slope by finding two points and using the slope formula. You can identify the $y-$ intercept by sight by finding where the line crosses the $y-$ axis on the graph. The value of $b$ is the $y-$ coordinate of this point.

Example 1

Write an equation in slope-intercept form that represents the following line.

First find the slope of the line. You already know how to do this using the slope formula. In this scenario, pick two points on the line to complete the formula. Use $(0,3)$ and $(2,2)$ .

$\text{slope} & = \frac{(y_2 - y_1)} {(x_2 - x_1)}\\ \text{slope} & = \frac{(2 - 3)} {(2 - 0)}\\ \text{slope} & = -\frac{1} {2}$

The slope of the line is $- \frac{1} {2}$ . This value will replace $m$ in the slope-intercept equation. Now you need to find the $y-$ intercept. Identify on the graph where the line intersects the $y-$ axis. It crosses the axes at $(0,3),$ so the $y-$ intercept is $3.$ This will replace $b$ in the slope-intercept equation, so now you have all the information you need to write the full equation. The equation for the line shown in the graph is $y = - \frac{1} {2} x + 3$ .

Equations of Parallel Lines

You studied parallel lines and their graphical relationships in the last lesson. In this lesson, you will learn how to easily identify equations of parallel lines. It’s simple—look for equations that have the same slope. As long as the $y-$ intercepts are not the same and the slopes are equal, the lines are parallel. (If the $y-$ intercept and the slope are the same, then the two equations would be for the same line, and a line cannot be parallel to itself.)

Example 2

Millicent drew the line below.

Which of the following equations could represent a line parallel to the one Millicent drew?

: D. $y = 2x + 1$
: C. $y = - 2x - 18$
: B. $y = \frac{1} {2} x + 9$
: A. $y = -\frac{1} {2} x - 6$

All you really need to do to solve this problem is identify the slope of the line in Millicent’s graph. Identify two points on the graph, and find the slope using the slope formula. Use points $(0,5)$ and $(1,3).$

$\text{slope} & = \frac{(y_2 - y_1)} {(x_2 - x_1)}\\ \text{slope}& = \frac{(3 - 5)} {(1 - 0)}\\ \text{slope}& = -\frac{2} {1}\\ \text{slope}& = -2$

The slope of Millicent’s line is $-2.$ All you have to do is identify which equation among the four choices has a slope of $-2.$ You can disregard all other information. The only equation that has a slope of $-2$ is choice C, so it is the correct answer.

Equations of Perpendicular Lines

You also studied perpendicular lines and their graphical relationships in the last lesson. Remember that the slopes of perpendicular lines are opposite reciprocals. In this lesson, you will learn how to easily identify equations of perpendicular lines. Look for equations that have the slopes that are opposite reciprocals of each other. In this case it doesn’t matter what the $y-$ intercept is; as long as the slopes are opposite reciprocals, the lines are perpendicular.

Example 3

Kieran drew the line in this graph.

Which of the following equations could represent a line perpendicular to the one Kieran drew?

: D. $y = -\frac{3} {2} x + 6$
: C. $y = -\frac{2} {3} x - 1$
: B. $y = \frac{2} {3} x - 4$
: A. $y = \frac{3} {2} x + 10$

All you really need to do to solve this problem is identify the slope of the line in Kieran’s graph and find its opposite reciprocal. To begin, identify two points on the graph, and find the slope using the slope formula. Use points $(0,2)$ and $(3,4)$ .

$\text{slope} & = \frac{(y_2 - y_1)} {(x_2 - x_1)}\\ \text{slope} & = \frac{(4 - 2)} {(3 - 0)}\\ \text{slope} & = \frac{2} {3}$

The slope of Millicent’s line is $\frac{2} {3}$ . Now find the opposite reciprocal of this value. The reciprocal of $\frac{2} {3}$ is $\frac{3} {2}$ , and the opposite of $\frac{3} {2}$ is $-\frac{3} {2}$ . So, $-\frac{3} {2}$ is the opposite reciprocal of $\frac{2} {3}$ . Now find the equation that has a slope of $-\frac{3} {2}$ . The only equation that has a slope of $-\frac{3} {2}$ is choice D, so it is the correct answer.

Equations in Standard Form

There are other ways to write the equation of a line besides the slope intercept form. One alternative is standard form. Standard form is represented by the equation below.

$Ax + By = C$

In this equation, both $A$ and $B$ cannot be $0$ . Also, if possible, $A$ and $B$ should be integers.

Example 4

Convert the equation $y = -\frac{1} {3} x + \frac{5} {7}$ into standard form.

The goal is to remove the fractions and have $x$ and $y$ on the same side of the equals sign. To start, multiply the entire equation by 7 to eliminate the denominator of $\frac{5} {7}$ .

$7y = -\frac{7} {3} x + 5$

Next multiply the equation by $3$ to eliminate the denominator of $-\frac{7} {3}$ .

$21y = -7x + 15$

Now add $7x$ to both sides of the equation to get $x$ and $y$ on the same side.

$21y + 7x & = -7x + 15 + 7x\\ 7x + 21y & = 15$

We are done. The equation in standard form is $7x + 21y = 15$ .

Lesson Summary

In this lesson, we explored how to understand equations of lines. Specifically, we have learned:

How to identify and write equations in slope-intercept form.
How to identify equations of parallel lines.
How to identify equations of perpendicular lines.
How to identify and write equations in standard form.

Always be on the lookout for ways to apply your knowledge of slope, parallel and perpendicular lines, and graphing on the coordinate plane to mathematical situations. Many problems in geometry can be solved by representing a geometric situation in the coordinate plane.

Points To Consider

Now that you understand equations of lines, you are going to take a closer look at perpendicular lines and their properties.

Last modified: Monday, June 28, 2010, 1:55 PM