Algebra

Forms of Linear Equation

Introduction

The equations which have the variables in the first order are termed linear equations, in other words in linear equations, the maximum value of the exponent of a variable is one. The linear equations can be of a single variable (e.g., x or y), and/or two or more variables, i.e., multivariable (e.g., x and y). Since an equation is a comparison of the mathematical algebraic expression by the equal sign, linear equations compare two algebraic expressions of the first order with an equal sign. In geometry, linear equations represent the equation of a straight line.

In real-life, conversion of any practical problem into the form of linear equations help us to solve them simply and accurately. In this process of formulation of the linear equations, one needs to be careful in the assignment of the known and unknowns in the formed linear equation.

The Standard Form of the Linear Equation

The standard form of the linear equation in the single variable can be expressed as,

${\rm{ax + b = 0}}$

where ${\rm{a,}}\;{\rm{and b}}$ are the constants that can have the real values except ${\rm{a}} \ne {\rm{0}}$ and ${\rm{x}}$ is the variable of the equation.

The standard form of the linear equation in the two variables can be expressed as,

${\rm{ax + by + c = 0}}$

where ${\rm{a,}}\;{\rm{b, and c}}$ are the constants that can have the real values except ${\rm{a, b}} \ne {\rm{0}}$ and ${\rm{x}}\;{\rm{and y}}$ are the variables of the equation.

Other Forms of the Linear Equations

Slope-intercept form

The slope-intercept form of the linear equation, which is also the most popular form, can be expressed as,

${\rm{y = mx + c}}$

where ${\rm{m}}$ is the slope of the equation and ${\rm{c}}$ is the intercept at the y-axis.

For the given equation in the slope-intercept form, the intercept of the x-axis can be defined as the,

${\rm{x = - }}\frac{{\rm{c}}}{{\rm{m}}}$

Which is obtained by substituting the value of ${\rm{y = 0}}$ in the above equation.

The slope of any straight line can be defined as the ratio of the rise and run (see below Fig.1), and therefore, can be defined as,

${\rm{m = }}\frac{{{\rm{rise}}}}{{{\rm{run}}}} = \frac{{{{\rm{y}}_{\rm{2}}} - {{\rm{y}}_{\rm{1}}}}}{{{{\rm{x}}_{\rm{2}}} - {{\rm{x}}_{\rm{1}}}}}$

Figure 1: The slope m of any equation by the concept of rise and run

Figure 2: Slope-intercept form of the linear equation having slope m and y-intercept c.

In the slope-intercept form the slope of the line parallel to the x-axis becomes zero, since for the change in the run there is no change in rise, whereas the slope of the line parallel to the y-axis becomes indeterministic since the denominator of the ratio of rise versus run becomes zero and makes m indeterministic.

Point-slope form

When the equation of the line is written in terms of the slope ${\rm{m}}$ and a point $\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}\;{{\rm{y}}_{\rm{1}}}} \right)$ , through which the straight line is passing, using the slope equation as defined above. We can write the equation of the line,

${\rm{y - }}{{\rm{y}}_1}{\rm{ = m}}\left( {{\rm{x - }}{{\rm{x}}_{\rm{1}}}} \right)$

This form of the linear equation can be also understood in the terms of the fact that the difference between the y-coordinates of two points on the line is directly proportional to the difference between the corresponding x-coordinates.

Figure 3: Point-slope form of the linear equation for the slope m and a point $\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}\;{{\rm{y}}_{\rm{1}}}} \right)$ on the straight line.

Two-point form

The two-point form of the linear equation is used to obtain the equation of the straight line when the two points it is passing through are given. Therefore, if we know two points $\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}\;{{\rm{y}}_{\rm{1}}}} \right)$ and $\left( {{{\rm{x}}_2}{\rm{,}}\;{{\rm{y}}_2}} \right)$ , the straight line is passing through then we can write the equation of the line such that,

Or we can also write,

${\rm{y - }}{{\rm{y}}_2}{\rm{ = }}\left( {\frac{{{{\rm{y}}_{\rm{2}}}{\rm{ - }}{{\rm{y}}_{\rm{1}}}}}{{{{\rm{x}}_{\rm{2}}}{\rm{ - }}{{\rm{x}}_{\rm{1}}}}}} \right)\left( {{\rm{x - }}{{\rm{x}}_2}} \right)$

Here, the variables in the equation $\left( {{\rm{x,}}\;{\rm{y}}} \right)$ show the random point on the straight line.

Figure 4: Two-point form of the linear equation passing through the points $\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}\;{{\rm{y}}_{\rm{1}}}} \right)$ and $\left( {{{\rm{x}}_2}{\rm{,}}\;{{\rm{y}}_2}} \right)$ .

Intercept form

The equation for a straight line in the intercept form is given by the formula,

$\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1$

Where a and b are the intercepts at the x- and y-axes, respectively.

Figure 5: Intercept form of the linear equation having a and b intercept on x- and y-axes, respectively.

Graphing the Linear Equations

There are certain features of the linear equations in the ${\rm{xy - }}$ plane that need to be utilised in the graphing of any linear equation. Since, any point on the linear equation is considered as a solution of the linear equation, therefore the points which are on the given equation of the line satisfies the given linear equation, and a locus of the points which satisfy the given equation of line are used to graph any given equation.

Step 1: Identify the given linear equation and for the given values of x determine the values of y using the given equation, where a point where x=0 and y=0, needs to be considered as well.

Step 2: Mark all these coordinates on the graph paper.

Step 3:The line joining all these coordinates is the graph of the given line of equation.

Applications of the Linear Equations

There are several applications of the linear equations, in which the important one is their use in real world problems. The linear equations can be used to obtain the unknowns from the following scenarios,

In the age related problems, we can find the age of the particular person.
In coordinate geometry we can solve the problems using the linear equations.
In economics the profit, money and percentage based problems can be solved.
In the laws of motion, they can be used to estimate the speed, distance and time.

Conclusion

Thus, the linear equation is the equation that compares the two algebraic equations with an equal sign and has the maximum value of the exponent of any variable equal to 1. The linear equations can be single-variable or multivariable. These equations can be utilised to obtain the unknown variables in different real-life problems. The different types of linear equations can be expressed as follows,

Slope-intercept form

${\rm{y = mx + c}}$ , where m is the slope of the line and c is the y-intercept.

Point-slope form

${\rm{y - }}{{\rm{y}}_1}{\rm{ = m}}\left( {{\rm{x - }}{{\rm{x}}_{\rm{1}}}} \right)$ , where m is the slope of the line and $\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}\;{{\rm{y}}_{\rm{1}}}} \right)$ is the point through which a straight line is passing.

Two-point form

${\rm{y - }}{{\rm{y}}_1}{\rm{ = }}\left( {\frac{{{{\rm{y}}_{\rm{2}}}{\rm{ - }}{{\rm{y}}_{\rm{1}}}}}{{{{\rm{x}}_{\rm{2}}}{\rm{ - }}{{\rm{x}}_{\rm{1}}}}}} \right)\left( {{\rm{x - }}{{\rm{x}}_{\rm{1}}}} \right)$ or ${\rm{y - }}{{\rm{y}}_2}{\rm{ = }}\left( {\frac{{{{\rm{y}}_{\rm{2}}}{\rm{ - }}{{\rm{y}}_{\rm{1}}}}}{{{{\rm{x}}_{\rm{2}}}{\rm{ - }}{{\rm{x}}_{\rm{1}}}}}} \right)\left( {{\rm{x - }}{{\rm{x}}_2}} \right)$

Where the line is passing through the points $\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}\;{{\rm{y}}_{\rm{1}}}} \right)$ and $\left( {{{\rm{x}}_2}{\rm{,}}\;{{\rm{y}}_2}} \right)$ .

Intercept form

$\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1$

Where a and b are the intercepts at the x- and y-axes, respectively.

Solved Examples

Q.1. A line passing through the points (2, 4) and having a slope of -2, then find the line of the equation?

Solution:

Given:

$\begin{array}{l}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}\;{{\rm{y}}_{\rm{1}}}} \right) = \left( {2,\;4} \right)\\{\rm{m}} = - 2\end{array}$

The given equation of line is a point slope form, since a point from which the line is passing through and slope of the line is given, therefore, we use the standard form of the point slope form to obtain the line of the equation, such that,

${\rm{y - }}{{\rm{y}}_1}{\rm{ = m}}\left( {{\rm{x - }}{{\rm{x}}_{\rm{1}}}} \right)$

$\begin{array}{l}{\rm{y - 4 = - 2}}\left( {{\rm{x - 2}}} \right)\\{\rm{y}} - 4 = - 2{\rm{x}} + 4\\2{\rm{x}} + {\rm{y = 8}}\end{array}$

Thus the line of the equation is $2{\rm{x}} + {\rm{y = 8}}$ .

Q.2. Determine the line of the equation having a slope of -5 and y-intercept of the equation is 3.

Solution:

Given:

$\begin{array}{l}{\rm{c}} = 3\\{\rm{m}} = - 5\end{array}$

The given equation of line is a slope-intercept form, since the slope of the line and its y-intercept is given, therefore, we use the standard form of the slope-intercept form to obtain the line of the equation, such that,

${\rm{y = mx + c}}$

$\begin{array}{l}{\rm{y = - 5x + 3}}\\{\rm{5x + y = 3}}\end{array}$

Thus the line of the equation is ${\rm{5x + y = 3}}$ .

Q.3. A line passing through the points (2, 4) and (5,3), then find the line of the equation?

Solution:

Given:

$\begin{array}{l}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}\;{{\rm{y}}_{\rm{1}}}} \right) = \left( {2,\;4} \right)\\\left( {{{\rm{x}}_2}{\rm{,}}\;{{\rm{y}}_{\rm{2}}}} \right) = \left( {5,\;3} \right)\end{array}$

The given equation of line is a two-point form, since two points from which the line is passing through are given, therefore, we use the standard form of the two-point form to obtain the line of the equation, such that,

${\rm{y - }}{{\rm{y}}_1}{\rm{ = }}\left( {\frac{{{{\rm{y}}_2} - {{\rm{y}}_1}}}{{{{\rm{x}}_2} - {{\rm{x}}_1}}}} \right)\left( {{\rm{x - }}{{\rm{x}}_{\rm{1}}}} \right)$

$\begin{array}{l}{\rm{y - 4 = }}\left( {\frac{{3 - 4}}{{5 - 2}}} \right)\left( {{\rm{x - 2}}} \right)\\{\rm{y}} - 4 = \frac{{ - 1}}{3}\left( {{\rm{x - 2}}} \right)\\3{\rm{y}} - 12 = - {\rm{x}} + 2\\{\rm{x}} + 3{\rm{y}} = 14\end{array}$

Thus the line of the equation is ${\rm{x}} + 3{\rm{y}} = 14$ .

Q.4. Determine the line of the equation having the x- and y-intercepts are 5 and 10, respectively.

Solution:

Given:

$\begin{array}{l}{\rm{a}} = 5\\{\rm{b}} = 10\end{array}$

The given equation of line is a intercept form, since the its x- and y-intercepts are given, therefore, we use the standard form of the intercept form to obtain the line of the equation, such that,

$\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1$

$\begin{array}{l}\frac{{\rm{x}}}{5} + \frac{{\rm{y}}}{{10}} = 1\\2{\rm{x + y = 10}}\end{array}$

Thus the line of the equation is $2{\rm{x + y = 10}}$ .

Q.5. A line passing through the points (0, 3) and (2,5), then find the line of the equation?

Solution:

Given:

$\begin{array}{l}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}\;{{\rm{y}}_{\rm{1}}}} \right) = \left( {0,\;3} \right)\\\left( {{{\rm{x}}_2}{\rm{,}}\;{{\rm{y}}_{\rm{2}}}} \right) = \left( {2,\;5} \right)\end{array}$

${\rm{y - }}{{\rm{y}}_1}{\rm{ = }}\left( {\frac{{{{\rm{y}}_2} - {{\rm{y}}_1}}}{{{{\rm{x}}_2} - {{\rm{x}}_1}}}} \right)\left( {{\rm{x - }}{{\rm{x}}_{\rm{1}}}} \right)$

$\begin{array}{l}{\rm{y - 3 = }}\left( {\frac{{5 - 3}}{{2 - 0}}} \right)\left( {{\rm{x - 0}}} \right)\\{\rm{y}} - 3 = \frac{2}{2}{\rm{x}}\\{\rm{y}} - 3 = {\rm{x}}\\{\rm{ - x}} + {\rm{y}} = 3\end{array}$

Thus the line of the equation is ${\rm{ - x}} + {\rm{y}} = 3$ .

FAQs

1. Name the different types of equations in mathematics.

Ans: An equation compares the two algebraic expressions by an equal sign. The different types of equations include the identities, which are truly independent of the value of the variables, however, the conditional equations are true only for the selective set of variables.

2. Name some important types of equations one observes and/or solves in mathematics.

Ans: Some important types of equations one usually observes and solves is,

Linear Equations
Radical Equations
Exponential Equations
Rational Equations

3. What is the most basic linear equation?

Ans: The most basic linear equation is defined in one variable and can be written as,

${\rm{ax}} + {\rm{b}} = 0$

Where the a and b are the real-valued constants with a condition that a cannot be equal to zero.

4. Name some of the methods used to solve the pair of linear equations.

Ans: Some of the methods that are used to solve the pair of linear equations are,

Substitution Method
Elimination Method
Graphing Method
Matrix Method

5. What are the different symbols used to express the different types of linear equations?

Ans: The different types of symbols that are used to express the linear equations use equal sign (=), and there are linear inequalities that use the symbols less than (<), more than (>), less than or equal to (≤), and/or more than or equal to (≥).

References

Forms of Linear Equations. (n.d.). Brilliant. Retrieved January 16, 2023, from https://brilliant.org/wiki/forms-of-linear-equations/
Forms Of Linear Equation (video lessons, examples, solutions). (n.d.). Online Math Learning. Retrieved January 16, 2023, from https://www.onlinemathlearning.com/forms-linear-equations.html
Home > Forms of Linear Equations – Explanation & Examples. (n.d.). The Story of Mathematics. Retrieved January 16, 2023, from https://www.storyofmathematics.com/forms-of-linear-equations/

Written by

Prerit Jain

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