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Coordinate Geometry Class 10 Notes CBSE Maths Chapter 7 (Free PDF Download)

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Class 10 Maths Revision Notes for Coordinate Geometry of Chapter 7 - Free PDF Download

CBSE Class 10 Coordinate Geometry Notes are proved to be excellent study material for students preparing for the board exams. These notes are prepared as per the guidelines issued by the CBSE board to help the students during their exam preparations. All the concepts of the chapter are explained in simple language in these notes by the expert teachers at Vedantu. CBSE Class 10 Coordinate Geometry Notes are available free of cost in PDF format for students’ ease.

The PDF covers all the important topics and the formulas required to solve the questions covered in each exercise of the chapter. So, download CBSE Class 10 Maths Notes Chapter 7 Coordinate Geometry to ease your exam preparation. Vedantu is a platform that provides free NCERT Solutions and other study materials for students. You can download Maths NCERT Solutions Class 10 and NCERT Solutions Class 10 Science to help you to revise the complete Syllabus and score more marks in your examinations.

Download CBSE Class 10 Maths Revision Notes 2024-25 PDF

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Coordinate Geometry Class 10 Notes CBSE Maths Chapter 7 (Free PDF Download)
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Important Topics Covered under CBSE Class 10 Maths Chapter 7 - Coordinate Geometry

The following is a list of all the important topics that are covered under the chapter on Coordinate Geometry:

  1. Equation of a Straight Line

  2. Distance Formula

  3. Distance from the Origin

  4. Section Formula

  5. Midpoint Formula

  6. Area of a Triangle

  7. Area of a Polygon

  8. The Centroid of a Triangle


Access Class 10 Maths Chapter 7 - Coordinate Geometry

Important Terms and Concepts:


Quadrants


  • “The coordinate axes are of perpendicular lines " \[XOX`\] " and " \[YOY`\] " intersecting at \[O\] .” 

  • The plane is divided into four quadrants by the axes.

  • The Cartesian plane is the plane that contains the axes.

  • The lines " \[XOX`\] " and " \[YOY`\] " are known as the  \[x\] -axis and  \[y\] -axis, respectively, and are commonly drawn horizontally and vertically as seen in the picture.

  • \[O\], is the Point of Intersection of the axes which is known as origin.

  • Abscissae are the values of \[x\] measured along the \[x\] -axis from \[O\]. \[x\] has positive values along \[OX\], but negative values along \[OX'\] .

  • Similarly, ordinate refers to the values of  \[y\] measured along the \[y\] axis from \[O\].

  • \[y\] has positive values along \[OY\], but negative values along \[OY'\].

  • The coordinates of a point are the ordered pair containing the abscissa and ordinate of a point.


Distance Formula:

  • To find the distance two points \[\text{A}\left( {{x}_{\text{1}}}\text{ , }{{\text{y}}_{\text{1}}} \right)\]and \[B\left( {{\text{x}}_{2}}\text{ , }{{\text{y}}_{2}} \right)\]


Distance Formula


  • From the Figure, 

$\text{AC = }{{\text{x}}_{\text{2}}}\text{- }{{\text{x}}_{\text{1}}}$

$\text{BC = }{{\text{y}}_{\text{2}}}\text{- }{{\text{y}}_{\text{1}}}$

Therefore, In \[\Delta ABC\],

By using the Pythagoras Theorem, we get

$\text{A}{{\text{B}}^{\text{2}}}\text{ = A}{{\text{C}}^{\text{2}}}\text{ + B}{{\text{C}}^{\text{2}}}$

$\text{A}{{\text{B}}^{\text{2}}}\text{ = }{{\left( {{\text{x}}_{\text{2}}}\text{- }{{\text{x}}_{\text{1}}} \right)}^{\text{2}}}\text{+ }{{\left( {{\text{y}}_{\text{2}}}\text{- }{{\text{y}}_{\text{1}}} \right)}^{\text{2}}}$

$\text{AB  = }\sqrt{{{\left( {{\text{x}}_{\text{2}}}\text{- }{{\text{x}}_{\text{1}}} \right)}^{\text{2}}}\text{+ }{{\left( {{\text{y}}_{\text{2}}}\text{- }{{\text{y}}_{\text{1}}} \right)}^{\text{2}}}}$


Section Formula:

  • To find the coordinates of a point which divides the line segment joining two given points in a given ratio (internally).


Section formula


  • Let \[P\left( x\text{ , y} \right)\]divide the join of \[\text{A}\left( {{x}_{\text{1}}}\text{ , }{{\text{y}}_{\text{1}}} \right)\] and \[B\left( {{\text{x}}_{2}}\text{ , }{{\text{y}}_{2}} \right)\] in the Ratio \[\text{m:n}\].

Here,


$\angle \text{ PAC = }\angle \text{ BPD}$


$\angle \text{ PCA = }\angle \text{ BDP}={{90}^{0}}$


Therefore, by using \[AA\] similarity method, we get,


$AC=x-{{\text{x}}_{1}}$


$PD={{\text{x}}_{2}}-x$


From Similarity Property, we get,


$\dfrac{\text{AC}}{\text{PD}}\text{ = }\dfrac{\text{m}}{\text{n}}$


$\text{        = }\dfrac{\text{x-}{{\text{x}}_{\text{1}}}}{{{\text{x}}_{\text{2}}}\text{-x}}$ 


$\text{        =}\dfrac{\text{m}}{\text{n}}$


Now, make \[x\] the subject of the formula, 


$\text{   nx - n}{{\text{x}}_{\text{1}}}\text{ = m}{{\text{x}}_{\text{2}}}\text{ - mx}$


$\text{  Mx + nx = m}{{\text{x}}_{\text{2}}}\text{ + n}{{\text{x}}_{\text{1}}}$


$\text{X}\left( \text{m + n} \right)\text{ = m}{{\text{x}}_{\text{2}}}\text{ + n}{{\text{x}}_{\text{1}}}$


Therefore,


\[\text{X = }\dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{ + n}{{\text{x}}_{\text{1}}}}{\text{m + n}}\]


Similarly, we can show that


\[\text{Y = }\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{ + n}{{\text{y}}_{\text{1}}}}{\text{m + n}}\]


Thus, the Co-ordinates of \[P\] are \[\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{ + n}{{\text{x}}_{\text{1}}}}{\text{m + n}},\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{ + n}{{\text{y}}_{\text{1}}}}{\text{m + n}} \right)\] .


Mid−Point Formula:

  • If \[P\] is the mid−point of \[AB\], then \[m:n\] , 

Therefore, the ratio becomes \[1:1\]

And hence,

$\text{X = }\dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{ + n}{{\text{x}}_{\text{1}}}}{\text{m + n}}$ 

$\text{X = }\dfrac{{{\text{x}}_{\text{2}}}\text{ + }{{\text{x}}_{\text{1}}}}{\text{1 + 1}}$. 

$\text{X = }\dfrac{{{\text{x}}_{\text{2}}}\text{ + }{{\text{x}}_{\text{1}}}}{\text{2}}$

Similarly, we get

\[\text{Y = }\dfrac{{{\text{y}}_{1}}\text{ + }{{\text{y}}_{2}}}{\text{2}}\]

Thus, the Co-ordinate of \[P\]are \[\left( \dfrac{{{\text{x}}_{1}}\text{ + }{{\text{x}}_{2}}}{\text{2}},\dfrac{{{\text{y}}_{1}}\text{ + }{{\text{y}}_{2}}}{2} \right)\]

  • Note:

When the point \[P\] divides the line joining \[AB\]  in the ration \[m:n\]  externally then,


The Co-ordinates of \[P\] are \[\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{ - n}{{\text{x}}_{\text{1}}}}{\text{m - n}},\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{ - n}{{\text{y}}_{\text{1}}}}{\text{m - n}} \right)\]


The Centroid of a Triangle:

  • The centroid is the point of intersection of three medians. 

  • It is the point of intersection of a median

  • $\text{AG : GD}$  is \[2:1\]


Centroid of a Triangle


To Find the Coordinates of the Centroid of a Triangle:

Let the coordinates of the vertices of \[\vartriangle ABC\] be \[\text{A}\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{y}}_{\text{1}}} \right)\text{, B}\left( {{\text{x}}_{\text{2}}}\text{,}{{\text{y}}_{\text{2}}} \right)\] and \[\text{ C}\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)\]


Let \[\text{G}\left( \text{x,y} \right)\]be the centroid of the \[\vartriangle ABC\]


Here, \[\text{D}\] is the midpoint of \[\text{BC}\] , and hence


By applying the mid−point formula, we get


\[\text{a = }\dfrac{{{\text{x}}_{2}}\text{+}{{\text{x}}_{3}}}{\text{2}}\] and \[\text{b = }\dfrac{{{\text{y}}_{2}}\text{+}{{\text{y}}_{3}}}{\text{2}}\]


We know that,


In a \[2:1\] ratio, point \[G\] splits the median.


Co-ordinates of Centroid of Triangle


Therefore, by applying the section formula, we get,


$\text{X = }\dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{ + n}{{\text{x}}_{\text{1}}}}{\text{m + n}}$


$\text{X = }\dfrac{\text{2}\left( a \right)\text{ + 1}\left( {{\text{x}}_{\text{1}}} \right)}{\text{2 + 1}}$


$\text{X = }\dfrac{\text{2}\left( \dfrac{{{\text{x}}_{\text{2}}}\text{ + }{{\text{x}}_{3}}}{2} \right)\text{ + }{{\text{x}}_{\text{1}}}}{3}$


$\text{X = }\dfrac{{{\text{x}}_{\text{1}}}+{{\text{x}}_{\text{2}}}\text{ + }{{\text{x}}_{3}}\text{ }}{3}$


Similarly,


$Y=\dfrac{\text{2}\left( b \right)\text{ + 1}\left( {{y}_{\text{1}}} \right)}{\text{2 + 1}}$


$\text{Y = }\dfrac{\text{2}\left( \dfrac{{{y}_{\text{2}}}\text{ + }{{\text{y}}_{3}}}{2} \right)\text{ + }{{\text{y}}_{\text{1}}}}{3}$


$\text{Y = }\dfrac{{{y}_{1}}\text{ + }{{\text{y}}_{3}}\text{ + }{{\text{y}}_{3}}}{3}$


Therefore, Coordinates of the centroid are \[\left( \dfrac{{{\text{x}}_{1}}\text{ + }{{\text{x}}_{2}}+{{x}_{3}}}{3},\dfrac{{{\text{y}}_{1}}\text{ + }{{\text{y}}_{2}}+{{y}_{3}}}{3} \right)\]


Area of the Triangle:

To find the area of a triangle whose vertices are \[\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{y}}_{\text{1}}} \right)\text{, }\left( {{\text{x}}_{\text{2}}}\text{,}{{\text{y}}_{\text{2}}} \right)\] and \[\text{ }\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)\]


Area of the Triangle


Let, \[\text{A}\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{y}}_{\text{1}}} \right)\text{, B}\left( {{\text{x}}_{\text{2}}}\text{,}{{\text{y}}_{\text{2}}} \right)\] and \[\text{ C}\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)\]be the vertices of a triangle \[\vartriangle ABC\]


$\text{Area of }\vartriangle \text{ ABC }$


$\text{= Area of Trapezium ABML + Area of Trapezium ALNC}$


$\text{    - Area of Trapezium BMNC}$

 

$\text{= }\dfrac{1}{2}ML\left( MB+LA \right)+\dfrac{1}{2}LN\left( LA+NC \right)-\dfrac{1}{2}MN\left( MB+NC \right)$


$=\dfrac{1}{2}\left( {{x}_{1}}-{{x}_{2}} \right)\left( {{y}_{2}}+{{y}_{1}} \right)+\dfrac{1}{2}\left( {{x}_{3}}-{{x}_{1}} \right)\left( {{y}_{1}}+{{y}_{3}} \right)-\dfrac{1}{2}\left( {{x}_{3}}-{{x}_{2}} \right)\left( {{y}_{2}}+{{y}_{3}} \right)$ 


$=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$


Arrow Method:

It is to obtain the formula for the area of the triangle


$ \dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\\end{matrix} \right| $


$=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$


Note:

  1. If the points \[\text{A, B}\] and \[\text{C}\] we take in the anticlockwise direction, then the area will be positive. 

  2. If the points we take in a clockwise direction, the area will be negative.

  3. So we always take the absolute value of the area calculated.

Therefore, 

Area of triangle

$=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$.

  1. If the area of a triangle is zero, then the three points are collinear.


Benefits of Studying Vedantu’s CBSE Class 10 Revision Notes on Maths Chapter 7 - Coordinate Geometry 

The following are the benefits of studying Vedantu’s CBSE Class 10 Revision Notes on Maths Chapter 7 - Coordinate Geometry:

  • Vedantu’s revision notes are prepared by expert teachers who have significant experience in the respective subjects. 

  • The notes are aligned with the Class 10 CBSE syllabus and follow the CBSE guidelines strictly.

  • The revision notes are in a simple and easy-to-understand language that will help students seamlessly grab the concepts and retain the information for longer periods of time.

 

Conclusion

Our revision notes have been curated by experts who cater to students’ need for simple, easy-to-understand content that they can remember for a significant period of time to be able to reproduce in their exams. We hope that students make the best use of our revision notes for the chapter on Coordinate Geometry and score desirably in their Class 10 Maths exams and boards.

FAQs on Coordinate Geometry Class 10 Notes CBSE Maths Chapter 7 (Free PDF Download)

1. Is it beneficial to memorize all the formulae of Coordinate Geometry?

Yes, it is highly beneficial for the students to memorize all the formulas of coordinate geometry. Learning the Chapter 7 Maths Class 10 Coordinate Geometry formulas are very helpful for students to solve the questions of this chapter in the exams accurately and efficiently.

2. Is CBSE Class 10 Maths Coordinate Geometry scoring topic for students?

Coordinate Geometry is considered the most simple and scoring topic. Students can easily score good grades if they get well versed with all the topics of the chapter.

3. How to calculate the area of a triangle when coordinates at the three vertices of the triangle are given according to Revision Notes of Chapter 7 of Class 10 Maths? 

To calculate the area of a triangle with coordinates of vertices as  A (x1y1), B (x2y2), and  C (x3y3), you apply the formula = ½ [x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2 )]. If the coordinates are A (1, -1), B (-4, 6), and C (-3, -5), after applying the formula, the area = ½ [1(6+5) +(-4) (-5+1) + (-3) (-1 -6)] = 24 square units.

4. What do you understand by section formula according to Revision Notes of Chapter 7 of Class 10 Maths?

Section formula is the formula that is used to calculate the coordinates of a point that divides a line segment in a ratio. So, if point P (x,y) divides the line segment joining A (x1,y1) and B (x2,y2) into the ratio m1: m2 , then by using the section formula we calculate the coordinates of point P. The section formula = (m1 x2 + m2 x1 / m1+ m2, m1 y2 + m2 y1 / m1+ m2).

5. Is the optional exercise important for Coordinate Geometry of Class 10 Maths?

Exercise 7.4 has been added as an optional exercise in the CBSE Chapter 7 of Class 10 Maths. Exercise 7.4 being an optional exercise means that it is not significant for examination. Although to get the idea of the coordinate geometry chapter, all students must attempt to solve questions in these exercises. Practising will help you in brushing up on your knowledge and making sure that you are able to solve any question from this chapter. 

6. How to prepare for Chapter 7 of Class 10 Maths?

For an effective preparation of CBSE Chapter 7 of Class 10 Maths, you should first focus on completing the exercises from the NCERT textbook along with examples and then initiate solving questions from other prescribed books. You can use Vedantu’s NCERT Solutions for Chapter 7 of Class 10 Maths to get hold of fully solved sums. To revise, use Vedantu’s Revision Notes of Chapter 7 of Class 10 Maths for a summarized version of the chapter with all important examples and formulas. 

7. Is Class 10 Maths Chapter 7 scoring?

All the chapters including Chapter 7 (Coordinate Geometry) in Class 10 Maths are extremely scoring. The correct method of study and study material can help you in scoring high in your Class 10 Maths paper. To score full marks in questions from this chapter, you should thoroughly practice NCERT exercise questions and examples. Your exam preparation is incomplete without revision and for that, you should use the best Vedantu’s Revision Notes of Chapter 7 of Class 10 Maths available on Vedantu. These solutions are available at free of cost on the Vedantu website(vedantu.com) and mobile app as well.