Like us!

Have a Look!

WhatsApp Message

Share Link

Mid Point Formula

Coordinate Geometry I - Concepts

Distance Formula

Section Formula

Mid Point Formula

Trisection Points of Line

Class - 10th IMO Subjects

Concept Explanation

Mid-Point Formula

Coordinates of the mid-point M (x,y) of the segment AB are obtained by taking $m_1 = m_2$ in the section formula:

$x=frac{m_{2}x_{1}+m_{1}x_{2}}{m_{1}+m_{2}};and;y=frac{m_{2}y_{1}+m_{1}y_{2}}{m_{1}+m_{2}}$

$x=frac{m_{1}x_{1}+m_{1}x_{2}}{m_{1}+m_{1}};and;y=frac{m_{1}y_{1}+m_{1}y_{2}}{m_{1}+m_{1}}$

$x=frac{m_{1}(x_{1}+x_{2})}{2m_{1}};and;y=frac{m_{1}(y_{1}+y_{2})}{2m_{1}}$

$large x=frac{1}{2}(x_{1}+x_{2}),$ $large y=frac{1}{2}(y_{1}+y_{2}),$

Illustration: Find the distance of the point (1,2) from the mid-point of the line segment joining the points (6,8) and (2,4).

Solution: Let the points (1,2), (6,8), and (2,4) be denoted by A, B, and C respectively. Let M be the mid-point of BC. Coordinates of M, the mid-point of BC, are given by

$large left ( frac{6+2}{2},frac{8+4}{2} right )=(4,6)$

We have $large AM^{2}=(4-1)^{2}+(6-2)^{2}$

$large =3^{2}+4^{2}=9+16=25$

$large Rightarrow$ $large AM=sqrt{25}=5;units$

Illustration: Find the lengths of medians of the triangle with vertices A(2,2), B(0,2), and C(2,-4).

Solution: Let the coordinates of A, B, and C be (2, 2), (0,2), and (2, -4) respectively. Let D, E, F be the mid-points of BC, CA, and AB respectively.

Then the coordinates of D, E, F are given by

$large Dleft ( frac{0+2}{2},frac{2-4}{2} right )$ , $large large Eleft ( frac{2+2}{2},frac{2-4}{2} right )$ , $large Fleft ( frac{0+2}{2},frac{2+2}{2} right )$

or D (1, -1) E ( 2, -1) , F (1, 2)

$large therefore$ Length of median AD= $large sqrt{(1-2)^{2}+(-1-2)^{2}}=sqrt{10}$

Length of median BE= $large sqrt{(2-0)^{2}+(-1-2)^{2}}=sqrt{13}$

Length of median CF = $large sqrt{(1-2)^{2}+[2-(-4)]^{2}}=sqrt{37}$

$large therefore$ Lengths of medians are $large sqrt{10},sqrt{13},sqrt{37};;units$ .

Illustration: Three consecutive vertices of a parallelogram are A(1,2), B(1,0) C(4,0). Find the fourth vertex D.

Solution: Let coordinates of D be (x,y). As ABCD is a parallelogram, the diagonals AC and BD bisect each other. If M is the mid-point of AC, then coordinates of M are $large left ( frac{1+4}{2},frac{2+0}{2} right )=left ( frac{5}{2},1 right )$

As M is the mid-point of BD also, coordinates of M are $large left ( frac{1+x}{2},frac{0+y}{2} right )=left ( frac{1+x}{2},frac{y}{2} right ).$

We have $large left ( frac{1+x}{2},frac{y}{2} right )=left ( frac{5}{2},1 right ) Rightarrow frac{1+x}{2}= frac{5}{2};and;frac{y}{2}=1$

$large Rightarrow 1+x=5;and;y=2Rightarrow x=4,y=2$

$large therefore$ Coordinates of D are (4,2)

Coordinates of Centroid of Triangle

The centroid of a triangle is the point of concurrence of the medians of a triangle.

If G is the centroid of triangle ABC, then G divides the median AD in the ratio 2:1.

Let D be the mid-point of BC, the coordinates of D are $large left ( frac{x_{2}+x_{3}}{2},frac{y_{2}+y_{3}}{2} right ).$

As G(x,y) divides AD in the ratio 2 : 1

$large x=frac{1(x_{1})+2left ( frac{x_{2}+x_{3}}{2} right )}{2+1}=frac{x_{1}+x_{2}+x_{3}}{3}$

and $large y=frac{1(y_{1})+2left ( frac{y_{2}+y_{3}}{2} right )}{2+1}=frac{y_{1}+y_{2}+y_{3}}{3}$

Thus, the coordinates of the centroid are

$large left ( frac{x_{1}+x_{2}+x_{3}}{3},frac{y_{1}+y_{2}+y_{3}}{3} right )$

Illustration: Find the third vertex of the triangle whose two vertices are (-3,1) and (0,-2) and the centroid is the origin.

Solution: Let the two given vertices be A(-3,1) and B(0,-2). Let the third vertex be C(x,y) . Coordinates of the centroid of $large Delta ABC$ is given by

$large left ( frac{-3+0+x}{3},frac{1-2+y}{3} right )=left ( frac{-3+x}{3},frac{-1+y}{3} right )$ .

As the centroid of $large Delta ABC$ is given to be the origin,

we have $large left ( frac{-3+x}{3},frac{-1+y}{3} right )=(0,0)$

$large Rightarrow ;;-3+x=0,;;-1+y=0$

$large Rightarrow ;;x=3,;;y=1$

Thus, the coordinates of the third vertex are (3,1).

Sample Questions

(More Questions for each concept available in Login)

Question : 1

(3, 4-a) is the mid-point of the lines segment joining points (2, a) and (4, 2). What is the value of a ?
A. 1	B. 8	C. -4	D. 2
Right Option : D
View Explanation

Question : 2

(3, 4) is the mid-point of the lines segment joining points (2, a) and (4, 4). What is the value of a ?
A. 1	B. 8	C. -4	D. 4
Right Option : D
View Explanation

Question : 3

If $p=(frac{a}{3},4)$ is the mid point of the segment joining the points Q (-6,5) and R (-2,3) . Then the value of a is :
A. -4	B. -6	C. 12	D. -12
Right Option : D
View Explanation

Video Link - Have a look !!!

Language - English

Chapters

Real Numbers I

Real Numbers II

Pair of Linear Equations I

Pair of Linear Equations II

Polynomial I

Polynomials II

Quadratic Equations I

Quadratic Equations II

Introduction to Trigonometry I

Introduction to Trigonometry II

Arithmetic Progression I

Arithmetic Progression II

Triangles I

Triangles II

Some Applications of Trigonometry

Coordinate Geometry I

Coordinate Geometry II

Introduction to Trigonometry

Direct and Inverse Variation

Surface Area and Volume I

Surface Area and Volume II

Areas Related to Circles

Content / Category

IMO-Mathematics Olympiad

Central Government Service

ISO - Science Olympiads

State Government Services

Banking And Insurance

Railways & Metro Services

MBA Entrance

Law Entrance

English Proficiency

Reasoning Proficiency

Computer Proficiency

Navodaya Entrance 6th & 9th

Sainik School Entrance

Class / Course

Students / Parents Reviews [10]

My experience was very good with Abhyas academy. I am studying here from 6th class and I am satisfied by its results in my life. I improved a lot here ahead of school syllabus.

Ayan Ghosh

8th

It was a good experience with Abhyas Academy. I even faced problems in starting but slowly and steadily overcomed. Especially reasoning classes helped me a lot.

Cheshta

10th

Abhyas is a complete education Institute. Here extreme care is taken by teacher with the help of regular exam. Extra classes also conducted by the institute, if the student is weak.

Om Umang

10th

Abhyas Methodology is very good. It is based on according to student and each child manages accordingly to its properly. Methodology has improved the abilities of students to shine them in future.

Manish Kumar

10th

A marvelous experience with Abhyas. I am glad to share that my ward has achieved more than enough at the Ambala ABHYAS centre. Years have passed on and more and more he has gained. May the centre f...

Archit Segal

7th

About Abhyas metholodology the teachers are very nice and hardworking toward students.The Centre Head Mrs Anu Sethi is also a brilliant teacher.Abhyas has taught me how to overcome problems and has...

Shreya Shrivastava

8th

Being a parent, I saw my daughter improvement in her studies by seeing a good result in all day to day compititive exam TMO, NSO, IEO etc and as well as studies. I have got a fruitful result from m...

Prisha Gupta

8th

I have spent a wonderful time in Abhyas academy. It has made my reasoning more apt, English more stronger and Maths an interesting subject for me. It has given me a habbit of self studying

Yatharthi Sharma

10th

My experience with Abhyas is very good. I have learnt many things here like vedic maths and reasoning also. Teachers here first take our doubts and then there are assignments to verify our weak poi...

Shivam Rana

7th

My experience with Abhyas academy is very good. I did not think that my every subject coming here will be so strong. The main thing is that the online tests had made me learn here more things.

Hiya Gupta

8th

View All

Attention !!!!!

Mid Point Formula

Mid-Point Formula

Coordinates of Centroid of Triangle

Students / Parents Reviews [10]

Ayan Ghosh

Cheshta

Om Umang

Manish Kumar

Archit Segal

Shreya Shrivastava

Prisha Gupta

Yatharthi Sharma

Shivam Rana

Hiya Gupta