Courses
Courses for Kids
Free study material
Free LIVE classes
More

# JEE Important Chapter - Coordinate Geometry

Get interactive courses taught by top teachers

## Coordinate Geometry - Introduction

Last updated date: 28th Mar 2023
Total views: 96.9k
Views today: 10

The study of geometric figures using coordinate axes is known as coordinate geometry. Straight lines, curves, circles, ellipses, hyperbolas, and polygons can all be conveniently drawn and represented to scale in the coordinate axes. Furthermore, using the coordinate system to work algebraically and examine the characteristics of geometric figures is referred to as coordinate geometry.

In this article, We will go through some of the important concepts of co-ordinate geometry like what is coordinate geometry - definitions, explanations and formulas. Also, we will go through some of the solved coordinate geometry problems for a better understanding of the concept.

## JEE Main Maths Chapter-wise Solutions 2022-23

### Important Topics of Coordinate Geometry

• Cartesian Coordinates

• Distance Formula

• Distance Between Two Points

• Slope

• Midpoint formula

• Equation of a line

• Slope-Intercept Form of a Line

• Point Slope Form

• Euclidean Distance Formula

### What is Coordinate Geometry?

Coordinate geometry is a field of mathematics that aids in the presentation of geometric shapes on a two-dimensional plane and to study their properties. To get a basic understanding of Coordinate Geometry, we will learn about the coordinate plane and the coordinates of a point initially.

### Coordinate Plane

A cartesian plane divides plane space into two dimensions, making it easier to locate points. The coordinate plane is another name for it. The horizontal X-axis and the vertical Y-axis are the two axes of the coordinate plane. The origin is the place where these coordinate axes connect, dividing the plane into four quadrants (0, 0). Furthermore, any point in the coordinate plane is represented by a point (x, y), where the x value represents the point's position relative to the X-axis and the y value represents the point's position relative to the Y-axis.

The point represented in the four quadrants of the coordinate plane has the following properties:

• The origin ‘O’ is the point of intersection of the X-axis and the Y-axis and has the coordinate points (0, 0) generally.

• The X-axis to the right of the origin ‘O’ is the positive X-axis and to the left of the origin, ‘O’ is the -(ve) X-axis. Also, the Y-axis above the origin ‘O’ is the +(ve) Y-axis, and below the origin ‘O’ is the -(ve) Y-axis.

• The point present in the first quadrant (x, y) has both +(ve) values and is plotted with reference to the +(ve) X-axis and the +(ve) Y-axis.

• The point represented in the second quadrant is (-x, y) is plotted with reference to the -(ve) X-axis and +(ve) Y-axis.

• The point represented in the third quadrant (-x, -y) is plotted with reference to the -(ve) X-axis and -(ve) Y-axis.

• The point represented in the fourth quadrant (x, -y) is plotted with reference to the +(ve) X-axis and -(ve) Y-axis.

### Coordinate Geometry Formulas

Coordinate geometry formulae make it easier to prove the various properties of lines and figures represented by coordinate axes. The distance formula, slope formula, midpoint formula, section formula, and line equation are included in the list of coordinate geometry formulas. In the below paragraph, we'll learn more about each of the formulas.

### Coordinate Geometry Distance Formula

The distance between two points $\left({x}_{1},{y}_{1}\right)$$(x_1, y_1)$ and $\left({x}_{2},{y}_{2}\right)$$(x_2, y_2)$ is equal to the square root of the sum of the squares of the difference between the X-coordinates and the Y-coordinates of the two given points. The formula for the distance between two points is as given below:

D = $\sqrt{\left({x}_{2}-{x}_{1}{\right)}^{2}+\left({y}_{2}-{y}_{1}{\right)}^{2}}$$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

### Slope Formula

The inclination of a line is measured by its slope. The slope can be determined by selecting any two locations on the line and measuring the angle formed by the line with the positive X-axis. m = tanθ is the slope of a line that is inclined at an angle 'θ' with the positive X-axis.  The slope of a line joining the two points $\left({x}_{1},{y}_{1}\right)$$(x_1, y_1)$ and ${x}_{2},{y}_{2}\right)$$x_2, y_2)$ is equal to m = $\frac{\left({y}_{2}-{y}_{1}\right)}{\left({x}_{2}-{x}_{1}\right)}$$\frac {(y_2 - y_1)}{(x_2 - x_1)}$.

m = tanθ

m = $\left({y}_{2}-{y}_{1}\right)$$(y_2 - y_1)$/$\left({x}_{2}-{x}_{1}\right)$$(x_2 - x_1)$

### Mid-Point Formula

The formula for finding the midpoint of the line connecting the points $\left({x}_{1},{y}_{1}\right)$$(x_1, y_1)$ and ${x}_{2},{y}_{2}\right)$$x_2, y_2)$ is a new point, whose abscissa is the average of the x values of the two given points, and the ordinate is the average of the y values of the two given points. The midpoint is on the line that connects the two points and is right in the middle of them.

$\left(x,y\right)=\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)$$(x, y) =\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$

### Section Formula in Coordinate Geometry

The section formula is useful to find the coordinates of a point that divides the line segment joining the points $\left({x}_{1},{y}_{1}\right)$$(x_1, y_1)$ and $\left({x}_{2},{y}_{2}\right)$$(x_2, y_2)$ in the ratio $m:n$$m : n$. The point that divides the provided two points is located on the line that connects them and can be found either between the two points or outside the line segment between them.

$\left(x,y\right)=\left(\frac{m{x}_{2}+n{x}_{1}}{m+n},\frac{m{y}_{2}+n{y}_{1}}{m+n}\right)$$(x, y) = \left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\right)$

### The Centroid of a Triangle

The centroid of a triangle is the point of intersection of medians of a triangle. (Median is a line that joins the vertex of a triangle to the mid-point of the opposite side). The centroid of a triangle having its vertices A$\left({x}_{1},{y}_{1}\right)$$(x_1, y_1)$, B$\left({x}_{2},{y}_{2}\right)$$(x_2, y_2)$, and C$\left({x}_{3},{y}_{3}\right)$$(x_3, y_3)$ is obtained from the following formula.

$\left(x,y\right)=\left(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}\right)$$(x, y) = (\dfrac{x_1+ x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3})$

### Area of a Triangle Coordinate Geometry Formula

The area of a triangle having the vertices A$\left({x}_{1},{y}_{1}\right)$$(x_1, y_1)$, B$\left({x}_{2},{y}_{2}\right)$$(x_2, y_2)$, and C$\left({x}_{3},{y}_{3}\right)$$(x_3, y_3)$ is obtained from the following formula. This formula is used to find the area of a triangle for all types of triangles.

Area of a Triangle = $\frac{1}{2}|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)|$$\dfrac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

### How to Find Equation of a Line in Coordinate Geometry?

With the use of a simple linear equation, this line equation represents all of the points on the line, ax + by + c= 0 is the standard form of a line equation. There are several methods for determining a line's equation. The slope-intercept form of the equation of a line (y = mx + c) is another essential form of the equation of a line. The slope of the line is m, and the Y-intercept of the line is c. The equation of a line also includes other types of line equations, such as point-slope form, two-point form, intercept form, and normal form.

y = mx + c

### List of Formulae

Sl.no

Name of the Concept

Formulae

1.

Distance formula

$D=\sqrt{\left({x}_{2}-{x}_{1}{\right)}^{2}+\left({y}_{2}-{y}_{1}{\right)}^{2}}$$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

2.

Slope, m of line ax+by+c=0

$m=-\frac{a}{b}$$m=-\dfrac{a}{b}$

3.

Angle between two lines is

$\theta ={\mathrm{tan}}^{-1}|\frac{{m}_{2}-{m}_{1}}{1-{m}_{2}{m}_{1}}|$$\theta = \tan^{-1}\left | \dfrac{m_2-m_1}{1-m_2m_1} \right |$

4.

Distance of a point p(x1,y1) from a line ax+by+c=0

$d=|\frac{a{x}_{1}+b{y}_{1}+c}{\sqrt{{a}^{2}+{b}^{2}}}|$$d=\left | \dfrac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \right |$

5.

Distance between two parallel lines of a slope m is

$d=|\frac{{c}_{1}-{c}_{2}}{\sqrt{1+{m}^{2}}}|$$d=\left | \dfrac{c_1-c_2}{\sqrt{1+m^2}} \right |$

### Solved Examples

Question 1. Find the equation of a line passing through (-2, 3) and having a slope of -1.

Solution:

The point on the line is $\left({x}_{1},{y}_{1}\right)=\left(-2,3\right)$$(x_1, y_1) = (-2, 3)$, and the slope is $m=-1$$m = -1$.

From the coordinate geometry point and slope form of the equation of the line, we get:

$\left(y-{y}_{1}\right)=m\left(x-{x}_{1}\right)\phantom{\rule{0ex}{0ex}}\left(y-3\right)=\left(-1\right)\left(x-\left(-2\right)\right)\phantom{\rule{0ex}{0ex}}y-3=-\left(x+2\right)\phantom{\rule{0ex}{0ex}}y-3=-x-2\phantom{\rule{0ex}{0ex}}x+y=3-2\phantom{\rule{0ex}{0ex}}x+y=1$$(y - y_1) = m(x - x_1) \\ (y - 3) =(-1)(x -(-2)) \\ y - 3 = -(x + 2) \\ y - 3 = -x -2 \\ x + y = 3 - 2 \\ x + y = 1$

Therefore the equation of the line is: x + y = 1

Question 2. Find the equation of a line having a slope of -2 and $y$$y$-intercept of 1.

Solution:

The given information is $m=-2$$m = -2$ and $y$$y$-intercept is $c=1$$c = 1$

From coordinate geometry we can make use of the slope intercept form of the equation of a line.

$y=mx+c\phantom{\rule{0ex}{0ex}}y=\left(-2\right)x+1\phantom{\rule{0ex}{0ex}}y=-2x+1\phantom{\rule{0ex}{0ex}}2x+y=1$$y = mx + c \\ y = (-2)x + 1 \\ y = -2x + 1 \\ 2x + y = 1$

Therefore the equation of the line we get is 2x + y = 1.

### Solved Problem of Previous year Question Paper

Question 1. The equations of two equal sides of an isosceles triangle are 7x − y + 3 = 0 and x + y − 3 = 0 and the third side passes through the point (1, 10). Then the equation of the third side is ___________.

Solution:

Any line passing through the point (1, 10) is given by y + 10 = m (x − 1)

Since it makes an equal angle say $\alpha$$\alpha$ with the given lines 7x − y + 3 = 0 and x + y − 3 = 0, therefore tan α = $\left[$$[$m − 7$\right]$$]$ / $\left[$$[$1 + 7m$\right]$$]$

= $\left[$$[$m − (−1)$\right]$$]$ / $\left[$$[$1 + m (−1)$\right]$$]$

⇒ m = $\left[$$[$1/3$\right]$$]$ or 3

So, the two possible equations of the third side are: 3x + y + 7 = 0 and x − 3y − 31 = 0.

Question 2. The locus of a point P, which divides the line joining (1, 0) and (2cosθ, 2sinθ) internally in the ratio 2:3 for all θ, is a ________.

Solution:

Let us consider the coordinates of the point P, that divides the line joining (1, 0) and (2 cosθ, 2sinθ) in the ratio 2:3 be (h, k). Then,

h = $\left[$$[$4 cosθ + 3$\right]$$]$/$\left[$$[$5$\right]$$]$ and k = $\left[$$[$4 sinθ$\right]$$]$/$\left[$$[$5$\right]$$]$

cosθ = $\left[$$[$5h − 3$\right]$$]$/$\left[$$[$4$\right]$$]$ and sinθ = $\left[$$[$5k$\right]$$]$/$\left[$$[$4$\right]$$]$

($\left[$$[$5h − 3$\right]$$]$/$\left[$$[$4$\right]$$]$)2 + ($\left[$$[$5k$\right]$$]$/$\left[$$[$4$\right]$$]$)2 = 1

(5h − 3)2 + (5k2) = 16

Therefore, locus of (h, k) is (5x − 3)2 + (5y)2 = 16, which is a circle.

Question 3. If the slope of a line passes through the point A (3, 2) is 3/4, then the points on the line which are 5 units away from A, are ___.

Solution:

The equation of line passes through the point (3, 2) and of slope 3/4 is 3x − 4y − 1 = 0

Let the point be (h, k) then,

3h−4k−1 = 0 …………..(i)     and

(h − 3)2 + (y − 2)2 = 52 (ii)

On solving the equations above, we get h = −1, 7 and k = −1, 5.

So, the points are (1, 1) and (7, 5).

### Practise Questions

Question 1. If the coordinates of vertices of a triangle are (0, 5), (1, 4) and (2, 5) then the coordinate of the circumcentre will be.

A.  (1, 5)

B.  $\left(\frac{3}{2},\phantom{\rule{mediummathspace}{0ex}}\frac{9}{2}\right)$$\left ( \dfrac{3}{2},\: \dfrac{9}{2} \right )$

C.  (1, 4)

D.  None of these

Question 2. The equation of the image of pair of rays y = |x| by the line x = 1 is

A.  |y| = x + 2

B.  |y| + 2 = x

C.  y = |x – 2|

D.  None of these

### Conclusion

Coordinate Geometry is recognized as one of the most fascinating mathematical concepts. Coordinate Geometry topics uses graphs with curves and lines to describe the relationship between geometry and algebra. Geometric aspects are provided in Algebra, allowing students to solve geometric problems. It is a type of geometry in which the coordinate points on a plane are expressed as an ordered pair of numbers. Here, the concepts of coordinate geometry are explained along with their formulas and solved coordinate geometry examples for better understanding.

See More

## JEE Main Important Dates

View all JEE Main Exam Dates
JEE Main 2023 January and April Session exam dates and revised schedule have been announced by the NTA. JEE Main 2023 January and April Session will now be conducted on 24-Jan-2023 to 31-Jan-2023 and 6-Apr-2023 to 12-Apr-2023, and the exam registration closes on 12-Jan-2023 and Apr-2023. You can check the complete schedule on our site. Furthermore, you can check JEE Main 2023 dates for application, admit card, exam, answer key, result, counselling, etc along with other relevant information.
See More
View all JEE Main Exam Dates

## JEE Main Information

Application Form
Eligibility Criteria
Reservation Policy
Exam Centres
NTA has announced the JEE Main 2023 January session application form release date on the official website https://jeemain.nta.nic.in/. JEE Main 2023 January and April session Application Form is available on the official website for online registration. Besides JEE Main 2023 January and April session application form release date, learn about the application process, steps to fill the form, how to submit, exam date sheet etc online. Check our website for more details. April Session's details will be updated soon by NTA.

## JEE Main Syllabus

View JEE Main Syllabus in Detail
It is crucial for the the engineering aspirants to know and download the JEE Main 2023 syllabus PDF for Maths, Physics and Chemistry. Check JEE Main 2023 syllabus here along with the best books and strategies to prepare for the entrance exam. Download the JEE Main 2023 syllabus consolidated as per the latest NTA guidelines from Vedantu for free.
See More
View JEE Main Syllabus in Detail

## JEE Main 2023 Study Material

View all study material for JEE Main
JEE Main 2023 Study Materials: Strengthen your fundamentals with exhaustive JEE Main Study Materials. It covers the entire JEE Main syllabus, DPP, PYP with ample objective and subjective solved problems. Free download of JEE Main study material for Physics, Chemistry and Maths are available on our website so that students can gear up their preparation for JEE Main exam 2023 with Vedantu right on time.
See More
All
Mathematics
Physics
Chemistry
See All

## JEE Main Question Papers

see all
Download JEE Main Question Papers & ​Answer Keys of 2022, 2021, 2020, 2019, 2018 and 2017 PDFs. JEE Main Question Paper are provided language-wise along with their answer keys. We also offer JEE Main Sample Question Papers with Answer Keys for Physics, Chemistry and Maths solved by our expert teachers on Vedantu. Downloading the JEE Main Sample Question Papers with solutions will help the engineering aspirants to score high marks in the JEE Main examinations.
See More

View all JEE Main Important Books
In order to prepare for JEE Main 2023, candidates should know the list of important books i.e. RD Sharma Solutions, NCERT Solutions, RS Aggarwal Solutions, HC Verma books and RS Aggarwal Solutions. They will find the high quality readymade solutions of these books on Vedantu. These books will help them in order to prepare well for the JEE Main 2023 exam so that they can grab the top rank in the all India entrance exam.
See More
Maths
NCERT Book for Class 12 Maths
Physics
NCERT Book for Class 12 Physics
Chemistry
NCERT Book for Class 12 Chemistry
Physics
H. C. Verma Solutions
Maths
R. D. Sharma Solutions
Maths
R.S. Aggarwal Solutions
See All

## JEE Main Mock Tests

View all mock tests
JEE Main 2023 free online mock test series for exam preparation are available on the Vedantu website for free download. Practising these mock test papers of Physics, Chemistry and Maths prepared by expert teachers at Vedantu will help you to boost your confidence to face the JEE Main 2023 examination without any worries. The JEE Main test series for Physics, Chemistry and Maths that is based on the latest syllabus of JEE Main and also the Previous Year Question Papers.
See More

## JEE Main 2023 Cut-Off

JEE Main Cut Off
NTA is responsible for the release of the JEE Main 2023 January and April Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2023 January and April Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general-category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.
See More

## JEE Main 2023 Results

NTA will release the JEE Main 2023 January and April sessions exam dates on the official website, i.e. {official-website}. Candidates can directly check the date sheet on the official website or https://jeemain.nta.nic.in/. JEE Main 2023 January and April sessions is expected to be held in February and May. Visit our website to keep updates of the respective important events of the national entrance exam.
See More
Rank List
Counselling
Cutoff
JEE Main 2023 state rank lists will be released by the state counselling committees for admissions to the 85% state quota and to all seats in IIT colleges. JEE Main 2023 state rank lists are based on the marks obtained in entrance exams. Candidates can check the JEE Main 2023 state rank list on the official website or on our site.

## JEE Top Colleges

View all JEE Main 2023 Top Colleges
Want to know which Engineering colleges in India accept the JEE Main 2023 scores for admission to Engineering? Find the list of Engineering colleges accepting JEE Main scores in India, compiled by Vedantu. There are 1622 Colleges that are accepting JEE Main. Also find more details on Fees, Ranking, Admission, and Placement.
See More

## FAQs on JEE Important Chapter - Coordinate Geometry

FAQ

1. What is a Cartesian Plane?

A cartesian plane in coordinate geometry is made up of two perpendicular lines called the X-axis (horizontal axis) and the Y-axis (vertical axis). The ordered pair can be used to calculate the exact position of a point in the Cartesian plane (x, y).

2. Where is Coordinate Geometry Used in Maths?

Coordinate geometry has a wide range of applications in mathematics. Many applications of coordinate geometry may be found in math topics such as vectors, three-dimensional geometry, equations, calculus, complex numbers, and functions. All of these topics require the presentation of data graphically in a two or three-dimensional coordinate plane.

3. What is the Application of Coordinate Geometry Used in Real Life?

In real life, coordinate geometry is used in a variety of ways. The coordinate system takes into account all of the maps we use to pinpoint locations, including Google Maps and physical maps. Furthermore, drawing land maps to scale is beneficial in large-scale land projects. To pinpoint any place in the seas, naval engineers employ coordinate systems.

## JEE Main Upcoming Dates

Vedantu offers free live Master Classes for CBSE Class 6 to 12, ICSE, JEE Main, JEE 2023, & more by India’s best teachers. Learn all the important concepts concisely along with amazing tricks to score high marks in your class and other competitive exams.
See More
JEE Main Exam April Session Starts
06 May 202334 days remaining
JEE Main Exam April Session Ends
12 May 202340 days remaining

JEE News
JEE Blogs