Lesson Explainer: Equation of a Straight Line: General Form Mathematics

In this explainer, we will learn how to find and write the equation of a straight line in general form.

We recall that the straight line with slope π‘š and 𝑦-intercept 𝑏 is described by the equation 𝑦=π‘šπ‘₯+𝑏.

This is the slope–intercept form of the equation of a straight line. There are many different ways to describe a line. For example, the equation of a line with slope π‘š passing through the point (π‘₯,𝑦) has the point–slope formπ‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯).

All the different ways to represent a line as an equation have something in common. A point (π‘₯,𝑦) lies on the line if and only if the equation of the line is satisfied when π‘₯=π‘₯ and 𝑦=π‘¦οŠ§. In this explainer, we introduce the general form of a straight line.

Definition: General Form of the Equation of a Straight Line

The general form of the equation of a straight line is given by π‘Žπ‘₯+𝑏𝑦+𝑐=0, where π‘Ž, 𝑏, and 𝑐 are real constants.

We remark that all lines can be written in the general form, while some equations of straight lines cannot be written in the point–slope or slope–intercept forms. For example, the line π‘₯=1 has no equation in slope–intercept form since the slope of this line is undefined. Nonetheless, the equation π‘₯=1 can be written in general form: π‘₯βˆ’1=0.

The general form of the equation of a straight line does not directly show the slope of the line. To obtain the slope of a line from its general form, it is beneficial to use the equation in slope–intercept form: 𝑦=π‘šπ‘₯+𝑏 where π‘š represents the slope of the line. We can convert an equation in its general form to one in the slope–intercept form by solving for the variable 𝑦.

Given the equation π‘Žπ‘₯+𝑏𝑦+𝑐=0 with 𝑏≠0, we subtract π‘Žπ‘₯ and 𝑐 from both sides to get 𝑏𝑦=βˆ’π‘Žπ‘₯βˆ’π‘.

Then, dividing both sides of the equation by 𝑏 leads to the slope–intercept form 𝑦=βˆ’π‘Žπ‘π‘₯βˆ’π‘π‘.

So, a line represented by the equation in its general form π‘Žπ‘₯+𝑏𝑦+𝑐=0, if 𝑏≠0, has slope βˆ’π‘Žπ‘ and 𝑦-intercept βˆ’π‘π‘.

Let us consider an example where we derive the slope of a line from the given equation in general form.

Example 1: Finding the Slope of a Line given Its Equation in General Form

A straight line has the equation βˆ’15π‘₯+3π‘¦βˆ’12=0. What is the slope of the line?

Answer

We are given the equation of a line in general form: βˆ’15π‘₯+3π‘¦βˆ’12=0.

To obtain the slope of the line, we should convert the above equation into the slope–intercept form 𝑦=π‘šπ‘₯+𝑏, where π‘š is the slope of the line and 𝑏 is the 𝑦-intercept. From the slope–intercept form, we can identify the slope, which is given by π‘š.

To convert the equation to the slope–intercept form, we should make 𝑦 the subject of the equation. Let us achieve this in steps. First, we add 15π‘₯ and 12 to both sides of the equation to get βˆ’15π‘₯+3π‘¦βˆ’12+15π‘₯+12=0+15π‘₯+12, which leads to 3𝑦=15π‘₯+12.

Next, we divide both sides by 3 to get 𝑦=5π‘₯+4.

So, we obtain the slope–intercept form 𝑦=5π‘₯+4. This tells us the slope of the line is 5.

The straight line with the equation βˆ’15π‘₯+3π‘¦βˆ’12=0 has a slope of 5.

In the next example, we obtain the π‘₯- and 𝑦-intercepts of a line from its equation in general form.

Example 2: Finding the π‘₯- and 𝑦-Intercepts of a Line

What are the π‘₯-intercept and 𝑦-intercept of the line 3π‘₯+2π‘¦βˆ’12=0?

Answer

Let us first consider the π‘₯-intercept of the line. We are given the equation of the line in general form: 3π‘₯+2π‘¦βˆ’12=0.

We recall that the π‘₯-intercept of a line has coordinates of the form (,0). So, the π‘₯-intercept of the line must satisfy the equation above with 𝑦=0. Substituting 𝑦=0 into the equation, we get 3π‘₯+2β‹…0βˆ’12=03π‘₯βˆ’12=0.

We then solve for π‘₯: 3π‘₯=12π‘₯=4.

Next, let us consider the 𝑦-intercept of the line. We recall that the 𝑦-intercept has coordinates of the form (0,), so the 𝑦-intercept must satisfy the equation 3π‘₯+2π‘¦βˆ’12=0 with π‘₯=0. Plugging π‘₯=0 into the equation gives us 3β‹…0+2π‘¦βˆ’12=02π‘¦βˆ’12=0.

We then solve for 𝑦: 2𝑦=12𝑦=6.

The line given by 3π‘₯+2π‘¦βˆ’12=0 has π‘₯-intercept 4 and 𝑦-intercept 6.

We note that the right-hand side of the equation of a line in general form is equal to zero. If we multiply both sides of the equation in general form by any nonzero constant, we would still end up with an equation in general form. For example, if a straight line is given in general form as 2π‘₯βˆ’π‘¦+3=0, then we can multiply through by 2 to get 4π‘₯βˆ’2𝑦+6=0.

So, 4π‘₯βˆ’2𝑦+6 is another equation of the same line in general form. This property is useful when simplifying an equation with rational coefficients. Say that we have the equation of a straight line in general form: 34π‘₯+π‘¦βˆ’5=0.

We can multiply both sides by 4 to cancel the denominator, 4: 34π‘₯Γ—4+𝑦×4βˆ’5Γ—4=0Γ—43π‘₯+4π‘¦βˆ’20=0.

Then, the equation of the same line in general form is more simply written as 3π‘₯+4π‘¦βˆ’20=0.

The general form of the equation of a line π‘Žπ‘₯+𝑏𝑦+𝑐=0 is closely related to its standard form: 𝐴π‘₯+𝐡𝑦=𝐢, where 𝐴, 𝐡, and 𝐢 are integers and 𝐴 is nonnegative. We can convert the standard form into general form by subtracting the constant 𝐢 from both sides of the equation.

Let us consider different examples where we derive the equation of a straight line in general form. In the next example, we find the general form of the equation of a line from its slope and 𝑦-intercept.

Example 3: Finding the Equation of a Line in General Form given Its Slope and 𝑦-Intercept

Write the equation of the line with slope 32 and 𝑦-intercept (0,3) in the form π‘Žπ‘₯+𝑏𝑦+𝑐=0.

Answer

We are given the slope 32 and 𝑦-intercept (0,3) of the line. We recall that the slope–intercept form for the equation of this line is 𝑦=32π‘₯+3.

We need to convert this equation into the form π‘Žπ‘₯+𝑏𝑦+𝑐=0, which is known as the general form of the equation of the line. We simplify the coefficients by multiplying both sides of the equation by 2 to get 2𝑦=3π‘₯+6.

We can subtract 2𝑦 from both sides of the equation to get 0=3π‘₯+6βˆ’2𝑦.

Then, we can rearrange the equation into the form π‘Žπ‘₯+𝑏𝑦+𝑐=0: 3π‘₯βˆ’2𝑦+6=0.

Hence, the equation of the line with slope 32 and 𝑦-intercept (0,3) can be written in general form as 3π‘₯βˆ’2𝑦+6=0.

When we are given two points (π‘₯,𝑦) and (π‘₯,𝑦) on the line, we recall that the slope of the line π‘š is given by π‘š==π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯.riserun

Then, using one of the points (π‘₯,𝑦), we can write the equation of the line in point–slope form: π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯).

We can use either of the points (π‘₯,𝑦) and (π‘₯,𝑦) to obtain the equation in point–slope form. Although the resulting equations will look different depending on the point chosen, they are both equivalent representations of the line. Often, though not always, either choice will lead to the same expression when converted to general form.

In the next example, we derive the equation of a straight line in general form given the coordinates of two points lying on the line.

Example 4: Finding the Equation of a Line through Two Points Giving the Answer in a Specified Form

Find the equation of the line that passes through the points 𝐴(βˆ’10,2) and 𝐡(0,5), giving your answer in the form π‘Žπ‘₯+𝑏𝑦+𝑐=0.

  1. 3π‘₯+10π‘¦βˆ’50=0
  2. βˆ’3π‘₯+10π‘¦βˆ’50=0
  3. βˆ’10π‘₯+3π‘¦βˆ’15=0
  4. 10π‘₯+3π‘¦βˆ’15=0
  5. 10π‘₯+3π‘¦βˆ’50=0

Answer

We recall that the slope of the line π‘š passing through points (π‘₯,𝑦) and (π‘₯,𝑦) can be computed by π‘š==π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯.riserun

We are given that the line passes through the points 𝐴(βˆ’10,2) and 𝐡(0,5). Letting 𝐴(βˆ’10,2) be (π‘₯,𝑦) and 𝐡(0,5) be (π‘₯,𝑦), we get π‘š=5βˆ’20βˆ’(βˆ’10)=310.

So, the slope of the line is 310.

To write the equation of this line, we also need a point on the line. We are given two points to choose from, and either choice is valid. For this problem, we notice that one of the points, 𝐡(0,5), is actually the 𝑦-intercept. So, choosing 𝐡 would lead to the slope–intercept form of the equation, while choosing 𝐴 would lead to a point–slope form of the equation. We will present both choices, and we will finish both methods by converting the resulting equations to the form π‘Žπ‘₯+𝑏𝑦+𝑐=0.

Method 1

We note that the point 𝐡(0,5) is on the 𝑦-axis, so it is the 𝑦-intercept of the line. Since we know the slope of the line π‘š=310, we can write the equation of the line in slope–intercept form: 𝑦=310π‘₯+5.

We need to convert this equation to the form π‘Žπ‘₯+𝑏𝑦+𝑐=0. Multiplying both sides of the equation by 10 gives us 10𝑦=3π‘₯+50.

Subtracting 10𝑦 from both sides of the equation, we get 0=3π‘₯+50βˆ’10𝑦.

Finally, rearranging the equation to the form π‘Žπ‘₯+𝑏𝑦+𝑐=0 gives us 3π‘₯βˆ’10𝑦+50=0.

In order to match one of the provided options, we multiply both sides of this equation by βˆ’1: βˆ’3π‘₯+10π‘¦βˆ’50=0.

Method 2

Let us use the point 𝐴(βˆ’10,2) and the slope 310 obtained above to write the equation of the line in point–slope form: π‘¦βˆ’2=310(π‘₯+10).

Multiplying both sides by 10 and distributing the parenthesis, we get 10π‘¦βˆ’20=3π‘₯+30.

Subtracting 3π‘₯ and 30 from both sides of the equation gives us βˆ’3π‘₯+10π‘¦βˆ’50=0.

We note that this equation matches one of the provided options.

So, the equation of our line is given in option B.

Given the π‘₯-intercept (𝐴,0) and the 𝑦-intercept (0,𝐡) of a line where 𝐴 and 𝐡 are nonzero, we can write the equation of the line in two-intercept form: π‘₯𝐴+𝑦𝐡=1.

We note that the π‘₯-intercept (𝐴,0) lies on the line since π‘₯=𝐴 and 𝑦=0, giving 𝐴𝐴+0𝐡=1.

Likewise, we can check for the 𝑦-intercept (0,𝐡) by substituting π‘₯=0 and 𝑦=𝐡, which gives us 0𝐴+𝐡𝐡=1.

This form is quite convenient since we do not need to compute the slope of the line to obtain it, and it is easy to sketch since we can mark both intercepts and connect them with a straight line as pictured below.

Since both expressions on the left-hand side of the equation in the two-intercept form are quotients, we begin the conversion process to its general form by multiplying both sides of the equation by the common denominator.

In the next example, we will derive the equation in general form when two intercepts are provided.

Example 5: Finding the Equation of a Straight Line

Determine the equation of the line that cuts the π‘₯-axis at 4 and the 𝑦-axis at 7.

Answer

Since the line cuts the π‘₯-axis at 4 and the 𝑦-axis at 7, the π‘₯- and 𝑦-intercepts of the line are 4 and 7, respectively. Let us write the equation in the general form: π‘Žπ‘₯+𝑏𝑦+𝑐=0. We present two different ways of approaching this problem. In the first method, we use the given intercepts to write the equation of the line in two-intercept form. In the second method, we use the intercepts to find the slope of the line, which is used to write the equation of the line in slope–intercept form.

Method 1

Since the π‘₯- and 𝑦-intercepts of the line are 4 and 7, we can write the equation of the line in two-intercept form: π‘₯4+𝑦7=1.

Multiplying both sides of the equation by the common denominator 28 gives us 7π‘₯+4𝑦=28.

By subtracting 28 from both sides of the equation, we obtain the general form 7π‘₯+4π‘¦βˆ’28=0.

Method 2

Since the π‘₯- and 𝑦-intercepts of the line are 4 and 7, we know that the line passes through the two points (4,0) and (0,7). We recall that the slope of the line is given by π‘š==7βˆ’00βˆ’4=βˆ’74.riserun

So, the slope of the line is βˆ’74. We also know that the 𝑦-intercept is 7. Then, we can write the equation of the line in slope–intercept form: 𝑦=βˆ’74π‘₯+7.

We can simplify this equation by multiplying both sides by 4. Then, 4𝑦=βˆ’7π‘₯+28.

Rearranging the equation gives us 7π‘₯+4π‘¦βˆ’28=0.

So, the equation of the line that cuts the π‘₯-axis at 4 and the 𝑦-axis at 7 is 7π‘₯+4π‘¦βˆ’28=0.

Key Points

  • The general form of the equation of a straight line is π‘Žπ‘₯+𝑏𝑦+𝑐=0, where π‘Ž, 𝑏, and 𝑐 are constants.
  • All straight lines can be represented by an equation in general form.
  • A line represented by an equation in its general form π‘Žπ‘₯+𝑏𝑦+𝑐, if 𝑏≠0, has slope βˆ’π‘Žπ‘ and 𝑦-intercept βˆ’π‘π‘. The equation of this line in its slope–intercept form is 𝑦=βˆ’π‘Žπ‘π‘₯βˆ’π‘π‘.
  • A line represented by the π‘₯- and 𝑦-intercepts in the form π‘₯𝐴+𝑦𝐡=1 can be represented in general form by multiplying the whole equation by the common denominator (π΄βˆ—π΅) and rearranging the terms: 𝐡π‘₯+π΄π‘¦βˆ’π΄π΅=0.

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