Skip to main content

Point-Slope Form: Definition, Examples, & Applications

Point-Slope Form: Definition, Examples, & Applications

In the field of mathematics and algebra, understanding the point-slope form is essential. The point-slope form plays a key role in graphing lines, especially in expressing a linear equation as well as unraveling how to find the slope of a given line.

The concept of the point-slope form empowers us to apprehend and solve problems related to linear equations. In point-slope form, a linear equation (equation of straight line) is represented with bi variables (two unknowns). As has been illustrated by the name, we require the gradient (slope) and a point on the straight line in order to construct an equation in the point-slope form.

In this blog, we will address the core concept of the point-slope form. We will elaborate on its definition, formula, and significant applications as well, and we will solve some examples in order to apprehend the core concept of the point-slope form in a precise and concise way.

Defining the Point-Slope Form?

The point-slope form is an important and fundamental concept that is used to represent the equation of a straight line. The point-slope form of a line is determined using the information about the slope-intercept form and the point from which the line passes (goes) through. It is often termed as the point-slope equation.

The Formula:

The point-slope form equation is written as:

y - y₁ = m (x - x₁)


·        y₁ is the y-coordinate (vertical axis) of the given point on the line.

·        x₁ is the x-coordinate (horizontal axis) of the same point.

·        m is the gradient of the line.

The point-slope form has a unique property which is its flexibility in its structure as compared to the other forms. This unique property enables you to use any point on a particular straight line to develop the point-slope form precisely and concisely.


To apply point-slope form, identify the specific point and slope relevant to your problem. Plug these values into the formula, and you'll have an equation that describes the relationship you're analyzing. Here we’ll solve some examples to apprehend the core concept of the point-slope form.

Example 1:

Determine the point-slope form if the slope of the line is -3 / 2 and it goes through the given point (7, -5).


Step 1: Given information

m = - 3, x = 7 and y = -5

Step 2: The general form of the point-slope form is given below:

y – y1 = m (x – x1)

Step 3: Put the values in the general form of the point-slope form of linear equations to get the required result.

y – (-5) = (-3 / 2) (x – 7)

y + 5 = -3/2 (x – 7) which is the point-slope form of a straight line.

Example 2:

Determine the point-slope form so that a straight line is passing through the points (1, -3) and (5, 7)


Step 1: Given data:

Here x1 = 1, x2 = 5,

y1 = - 3 and y2 = 7

Step 2: Now we will proceed to find the slope of the straight line by using the given formula: 

m = (y2 – y1) / (x2 – x1)

Step 3: Place the values to calculate the slope of the straight line.

m = (7 – (- 3) / (5 – 1)

m = (7 + 3) / 4

m = 10 / 4

m = 5 / 2

Step 4: The general form of the point-slope form is given below.

y – y1 = m (x – x1)

Here we can use any one of the given points on the line (x1, x2). Let us use (1, -3)

Step 5: Put the values in the general form of the point-slope form of the linear equations.

y – (- 3) = 5 / 2 (x – 1)

y + 3 = 5 / 2 (x – 1) which is the required point-slope form of a linear equation.

Example 3:

Find out the point-slope form so that a straight line having slope m = - 7 / 3 passes through the point (-2, -3).


Step 1: Given data:

m = - 7 / 3,

x1 = - 2 and y1 = - 3

Step 2: The general point-slope form of the linear equations is given below.

y – y1 = m (x – x1)

Step 3: Place the values in the above equation.

y – (- 3) = - 7 / 3 (x – (- 2))

y + 3 = - 7 / 3 (x + 2) which is the required equation of the point-slope form of a linear equation.

A point slope calculator make solving equations of lines quicker and easier instead of solving them manually by using point slope form.

Applications of the Point-Slope Form

The point-slope form isn't just a theoretical concept, it has real-world applications including:


In the field of engineering especially in the subject of civil engineering, the point-slope form is very useful and plays a key role in designing different sorts of structures and creating sketches and patterns for bridges, roads, and buildings.


In the field of physics, the point-slope form is very useful to signify the motion of objects like the path of the moving objects or their trajectory. 


Economists utilize point-slope form when analyzing supply and demand curves, helping to predict market trends.


Geographers use it to study gradients and slopes of terrain, aiding in cartography and mapmaking.


The point-slope form offers a straightforward way to express linear equations, making it easy to grasp and work with.


It's versatile, allowing you to quickly adapt equations to different scenarios by altering the point and slope.

Wrap Up:

The point-slope form is a powerful tool in algebra, offering a concise way to express linear equations. In this blog, we have discussed the concept of the point-slope form. We have defined the point-slope form, formula, and its important applications. In the last section, we have solved the examples of the point-slope form, equation of the straight line, etc.

By Anil Singh | Rating of this article (*****)

Popular posts from this blog

React | Encryption and Decryption Data/Text using CryptoJs

To encrypt and decrypt data, simply use encrypt () and decrypt () function from an instance of crypto-js. Node.js (Install) Requirements: 1.       Node.js 2.       npm (Node.js package manager) 3.       npm install crypto-js npm   install   crypto - js Usage - Step 1 - Import var   CryptoJS  =  require ( "crypto-js" ); Step 2 - Encrypt    // Encrypt    var   ciphertext  =  CryptoJS . AES . encrypt ( JSON . stringify ( data ),  'my-secret-key@123' ). toString (); Step 3 -Decrypt    // Decrypt    var   bytes  =  CryptoJS . AES . decrypt ( ciphertext ,  'my-secret-key@123' );    var   decryptedData  =  JSON . parse ( bytes . toString ( CryptoJS . enc . Utf8 )); As an Example,   import   React   from   'react' ; import   './App.css' ; //Including all libraries, for access to extra methods. var   CryptoJS  =  require ( "crypto-js" ); function   App () {    var   data

List of Countries, Nationalities and their Code In Excel File

Download JSON file for this List - Click on JSON file    Countries List, Nationalities and Code Excel ID Country Country Code Nationality Person 1 UNITED KINGDOM GB British a Briton 2 ARGENTINA AR Argentinian an Argentinian 3 AUSTRALIA AU Australian an Australian 4 BAHAMAS BS Bahamian a Bahamian 5 BELGIUM BE Belgian a Belgian 6 BRAZIL BR Brazilian a Brazilian 7 CANADA CA Canadian a Canadian 8 CHINA CN Chinese a Chinese 9 COLOMBIA CO Colombian a Colombian 10 CUBA CU Cuban a Cuban 11 DOMINICAN REPUBLIC DO Dominican a Dominican 12 ECUADOR EC Ecuadorean an Ecuadorean 13 EL SALVADOR

23 Best Key Features of MVC 6 and MVC 5

What’s new In MVC 6? The Added Key Features as following as, 1. The Microsoft makes a bundle of MVC, Web API, WebPages, SignalR, that bundle we called  MVC6 . 2. The MVC 6   added new cloud computing optimization system of MVC, web API, SignalR and entity framework. 3. In MVC 6, Microsoft removed the dependency of system.web.dll from MVC 6 because it's so expensive. Typically it consumes 30K memory per request/response. 4. Right now, in MVC 6 consume 2K memory per request response. It's too small memory consume. 5. Most of the problem solved using the  Roslyn Compiler . 6 . It’s added a  Start-up  class that replaces to  global.asax  file. 7. The Session state and caching adjusts our behavior depending on your hosting environment. 8. Host agnostic and its true side-by-side deployment 9. The vNext is a cross platform and open source and it's also supported to Mac, Linux, etc. 10. It’s also added to TagHeaplers use to creating an

Angular 7 and 8 Validate Two Dates - Start Date & End Date

In this example, I am sharing “ How to compare or validate two dates in Angular? ” using custom validation function in Angular 7 and Angular 8 . Here, I’m validating the two dates  - a start date and end date. The end date should be greater than the Start date”. Let’s see the example :- import { Component , OnInit } from '@angular/core' ; import { UserRequest } from '../model/user' ; @ Component ({   selector: 'User_Cal' ,   templateUrl: './usercal.component.html' ,   styleUrls: [ './usercal.component.css' ] }) export class UserCalComponent implements OnInit {   constructor ( private EncrDecr : EncrDecrService , private   http :  HttpClient ,               private datePipe : DatePipe ) {                            }   //model class   model = new UserRequest ( null , null , null , null , null );   //Error Display   error : any ={ isError: false , errorMessage: '' };   isValid

Encryption and Decryption Data/Password in Angular

You can use crypto.js to encrypt data. We have used 'crypto-js'.   Follow the below steps, Steps 1 –  Install CryptoJS using below NPM commands in your project directory npm install crypto-js --save npm install @types/crypto-js –save After installing both above commands it looks like  – NPM Command  1 ->   npm install crypto-js --save NPM Command  2 ->   npm install @types/crypto-js --save Steps 2  - Add the script path in “ angular.json ” file. "scripts" : [                "../node_modules/crypto-js/crypto-js.js"               ] Steps 3 –  Create a service class “ EncrDecrService ” for  encrypts and decrypts get/set methods . Import “ CryptoJS ” in the service for using  encrypt and decrypt get/set methods . import  {  Injectable  }  from   '@angular/core' ; import   *   as   CryptoJS   from   'crypto-js' ; @ Injectable ({    providedIn:   'root' }) export   class   EncrDecrS