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This chapter introduces you to the fundamental concepts of coordinate geometry, including Cartesian coordinates, distance formula, and section formula.
The chapter covers various concepts related to coordinate geometry, such as plotting points on a graph, finding the distance between two points, and dividing a line segment into two parts. The solutions provided by Swastik Classes offer step-by-step explanations to all the exercises and are designed to help you understand the concepts and solve problems effectively.
Coordinate geometry is an essential part of the mathematics curriculum and is widely used in various fields such as physics, engineering, and computer science. The solutions provided by Swastik Classes will help you to build a strong foundation in coordinate geometry and prepare you for more advanced mathematical concepts.
Overall, the NCERT Solutions for Class 10th Maths Chapter 7 – Coordinate Geometry, provided by Swastik Classes, are an excellent resource for students who want to excel in the subject. The solutions will not only help them score well in their exams but also prepare them for more advanced topics in mathematics. It is crucial to have a clear understanding of the concepts covered in this chapter as it forms the foundation for more advanced concepts in coordinate geometry.

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Answers of Physics NCERT solutions for Class 10th Maths Chapter 7 Coordinate Geometry

Unit 7

Coordinate Geometry

Exercise 7.1

Question 1:

Find the distance between the following pairs of points: 

(i)     (2, 3), (4, 1) 

(ii)     (−5, 7), (−1, 3) 

(iii)     (a, b), (− a, − b)

Answer:

  1. Distance between the two points is given by

dlUQOiHQKxzNk cbR8o d8YNAfem2h2D0fFvRKBmENzM1xg S5agWmfmawB7jInLtamm824ERSeWVvDL0aqDQ37cQAeCF5V ZYL7LKQjVMv1gWhjx

  1.     Distance between is given by T651PISOIDVEWFkxH6GiHG4ELhTdQSQKwJQpVrIZlnI4m1BGBewheQC3wjCB9t1Q6e58wNYaGrn 5m6nSuKKtUXm84iyo1QmlvwiCaovm2qOqCjuoydbp5H1G5gJ0ipmOShPk U
  1. Distance between and is given by

BZ5MILXIOaBymyziTLyEB04GfPSi

Question 2:

Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Answer:Wqi6U8NYqKcjFshjba5uqVvQpOd4rcFTF4Kl3UpvDiSwUUX2b0beunUgrJI2UOs9EeLoA9T6 HYUHsJPK98j9cJ4keYHpwHuOe3tUZirZ N BT43Hrv9VjVHHhRGtAEQr sHmXw

Distance between points
SDjjSST 2ta2KNoHO0pVxSad

Yes, we can find the distance between the given towns A and B.

Assume town A at origin point (0, 0).

Therefore, town B will be at point (36, 15) with respect to town A.

And hence, as calculated above, the distance between town A and B will be

39 km.

Question 3:

Determine if the points (1, 5), (2, 3) and (− 2, − 11) are collinear.

Answer:

Let the points (1, 5), (2, 3), and (−2, −11) be representing the vertices A, B, and C of the given triangle respectively.
eI2TMNYhzC3MRn7Is2gB4RjIG0s7YllFrN9BoAgmKqTDAFEYxW9xNWY0RPGedZt6HjiQKKvQ8jpqfnBj

Let
2hTA1pXL TIiVtFo622Cv0 S

Therefore, the points (1, 5), (2, 3), and (−2, −11) are not collinear.

Question 4:

Check whether (5, − 2), (6, 4) and (7, − 2) are the vertices of an isosceles triangle.

Answer:

Let the points (5, −2), (6, 4), and (7, −2) are representing the vertices A, B, and C of the given triangle respectively.
5bFKwX9TyBaNaVXUCrYYs1apCu8UZdlHEw7oQC FqfoykWa30nH mvYdXdAiMIPmc

As two sides are equal in length, therefore, ABC is an isosceles triangle.

Question 5:

In a classroom, 4 friends are seated at the points

 A, B, C and D as shown in the following figure. 

Champa and Chameli walk into the class and 

after observing for a few minutes Champa asks 

Chameli, “Don’t you think ABCD is a square?” 

Chameli disagrees.

Using distance formula, find which of them is correct.

Answer:

It can be observed that A (3, 4), B (6, 7), C (9, 4), and D (6, 1) are the positions of these 4 friends.

00NN04j2wzuCI4csGFR n dPttPKLzEGTjHOrvt3VE8e PRWhXKmvBepO7CC4Ip3mY6Tg 19n8vUyu XXif jrxSfzFb1VG9Uw7y2jaVYpNbRcyJH7TtdgkY qBIS70cZwGJWfI
21LdzTK mxd7Nky b4TiUS6e5enB mThuBXK4z6jIOgzVR6SChU6YpNFcXWyhlehzktEP 0geKz ZAOpowM0qmf g7U3wJc3Yp0Zst6XDEAKdV1 au2BwKb4TFtMQRbi2CWToNE
sbidkkm Ztz1l22AAD46IdQdWQgJ sI8PEjkZdHeAuSK9WnpiU3pov6ypHAXfggqYImddkMchD7LEVXnL2Ymb1QqSq4w2RH 6LxUOTZup PhY4KY1QORLNmg1bMaKIbvFo0 XFE
Hl28y2dFJFM21l7lAFqil4d8 lbp0j QuKZgYdkiN4vurs6knJoBE6P5QGVZdnaqjEbzHoohd1q0G Wf0uPJze4COSu aePj6Ku90AOfVJhRkroWKlhuTwMmaJ2nIFg YoP5Oe0
sk sOP0ifH5iugcjHk5lO9c473kMtLAyitgI7QilREapiRRLktvaI3g4CfVuzPj9lzZhpfdolugF46 nUakxYgnWCi

It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are of the same length.

Therefore, ABCD is a square and hence, Champa was correct.

Question 6:

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

(i)  (− 1, − 2), (1, 0), (− 1, 2), (− 3, 0)

(ii) (− 3, 5), (3, 1), (0, 3), (− 1, − 4)

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Answer:

  1.     Let the points (−1, −2), (1, 0), (−1, 2), and (−3, 0) be representing the vertices A, B,     C, and D of the given quadrilateral respectively.
YzkOvmj4qLO5YFx5v6uECuyw1I6sVNPIeDZF8WPAMAYBlhtKu1bfiNJUJ1SxVYcMjrdPJ3cbuRpnv2sGBLWveOqveIkdfI2ezZCz89Vmdw3BTiZZLel0zT z2N2MurGTFBgTl2E

It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square.

  1. Let the points (− 3, 5), (3, 1), (0, 3), and (−1, −4) be representing the vertices A, B,     C, and D of the given quadrilateral respectively.
TqPAP6B52TF62VLjn6zKT 5YmG2TVpMOhMuPmqrb0YcLSgO7fxnH 3WiRSBMvwZ0A0O yhfO38vCg7n3vmYf2PjVGYcyj qLo7MvfycGrV7qzd7D2omr7yuNfUPEMPQRl2k1HlU

It can be observed that all sides of this quadrilateral are of different lengths. Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc.

  1. Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C,     and D of the given quadrilateral respectively.

It can be observed that opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.

*Question 7:

Find the point on the x-axis which is equidistant from (2, − 5) and (− 2, 9).

Answer:

We have to find a point on x-axis. Therefore, its y-coordinate will be 0.

Let the point on x-axis be
lPYBs3Reikh4F7md48hND4HqLBy6SyEThzuUFoeKuf2n1TXmaU pP2EFveNHwX0Jk1JeULJpegeJ2p7YniIy0imoODRqdz5 0oiXNSRc7sZvpmJt

By the given condition, these distances are equal in measure.

ptCbDGR1beWJpE7Ak6qUER3mT3LIhzHEJAzD8ToELQCPtsYckDxRoWFGnKY6P5CowXSFKZ3L2AqP3yUX3P5aaVtLQGz0yKCzvxsPdDPDZkaGO4AjhIG RIgqWtewFQHUa FehUI

Therefore, the point is (− 7, 0).

*Question 8:

Find the values of y for which the distance between the points P (2, − 3) and Q (10, y) is 10 units.

Answer:

It is given that the distance between (2, −3) and (10, y) is 10.
jQ7aaBShWmkoUks6babIxm6t0XaEGxdaJnkqquVrU64jojbWwzfHBCtEQV4QBYUd4Pdwu O ZmZgJRrpfKH2tqkAfypeiDvwDOyIDnJ4uNeohEMyrq3YJR1RVq8nblU3WLcwEIg

*Question 9:

If Q (0, 1) is equidistant from P (5, − 3) and R (x, 6), find the values of x. Also, find the distance QR and PR.

Answer:

g7JkI03rYxcN2Trlb5uIsqoRo9rsbp8VR c90aF4huJrlF6Npa 6UfpvRABQXZwesnmEafZh9o1RP9cQxjwyuBFaX wM5pu 3FktHTE9h5bBOvDiTmZp903hy3PSSYRMHVKMG3A

Therefore, point R is (4, 6) or (−4, 6).

When point R is (4, 6),avzX1osKWDC0kZAr6u4KsKL5mHDbr2hOj9gTeB5Y7FXrdwRNRqAkIjgsVGDc

When point R is (−4, 6),tIVow5uCRKdJZgPos4l7iLf11kVJ5Z53M 8QA 3qoUD9g6hAgBhuTTiPGoNSnocN7TgxWNvPGLxqdEKaC8 VJonw3uY3MnVX3e NFc0J9h7S HKgnuVNx7eNAusH2 421vzvrgI

*Question 10:

Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (− 3, 4).

Answer:

Point (x, y) is equidistant from (3, 6) and (−3, 4).

BfCpKm6aYQaZamQFXUV

x-32+y-62=x+32+y-42

9-6x+36-12y=9+6x+16-8y

12x+4y=20

3x+y=5

Hence the relation between x and y can be given by: 3x+y=5

Exercise 7.2

Question1: 

Find the coordinates of the point which divides the join of (− 1, 7) and (4, − 3) in the ratio 2:3.

Answer:

Let P(x, y) be the required point. Using the section formula, we obtain

Therefore, the point is (1, 3).

Question 2:

Find the coordinates of the points of trisection of the line segment joining (4, − 1) and (− 2, − 3).

Answer:
BTsmuAV54WlwRKtg9hpuz0Xxc2KoiLDS8rMQ912emGngQVuM5qC4SakLUsoCeb4n3jHRP Ph9Vc6mYRoRHxR3nf7r5lFRUzcmY17w4AwhrFw7L6l93I6nDIRk nndIPJzd2WdLo

Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2.

yQV FZ4peAdbQmpfLIqDKVKl6pEY4JtBaMYuSDTxcdq6cmWDiJ82NBYZ4lt4WRxtQECjBx8 OHj9mhov6jxFW

Point Q divides AB internally in the ratio 2:1.
DYpxQGoPN0I0kq0IH726o8dpC5PMXrKV5XnKx1rFZ crGKocrhIpyWI3F8a5z8GXj7G0WqdD7rj5U2xwyhAzvdAJZjTAOhJPEAU 0X8htdCjlITy9nbM1BtDAG8DsSNdFz0 pc

Question 3:

To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following

figure. Niharika runs the distance AD on the 2nd line and posts a green flag. Preet runs the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
IJhWdRvNvu oyKz3raZdEhwO5qGiJMJa5xxylgpwH3bQbbHvgPyGNtmSt9pS50vVQlutVuQWgoaVZeaP XHcyMxDu7BQCX8lDToQqMNsA sWt r2gAmk7iVbC9FZSlwFYc a80

Answer: 

It can be observed that Niharika posted the green flag atof the distance AD i.e., point G is (2, 25). from the starting point of 2nd line. Therefore, the coordinates of this

Similarly, Preet posted red flag atof the distance AD i.e.,    m from the starting point of 8th line. Therefore, the coordinates of this point R are (8, 20).

Distance between these flags by using distance formula = GR

= idtncxWedUKACzD7s zBLAOoWTd4gNmshxz4cmY3VMMOO0zZpSmw

The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be A (x, y).

9qyMzs z7A5O3GKEKu4azvzLnqCFb 3ASxAlguVaWwzH6RpGY5d pY A7pMoJFpGH A2D2RJhQED3KN85vyDgeo9aeh n93DUDqr2hdeS1oBWF583UcHyyL5vzmdvh48IfhQ2Xg

Therefore, Rashmi should post her blue flag at 22.5m on 5th line.

Question 4:

Find the ratio in which the line segment joining the points (− 3, 10) and (6, − 8) is divided by (− 1, 6).

Answer:

Let the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by point (−1, 6) be k:1.
Hqtz2MpK0dges4N4mhHMw5JtdN1vGqS2Ax1SZFtJJtGqWgkLemlZHCilNvfpBMdvcoiYaK1NCaUgUBZpMXT9R4mnB ovnzJV4dqAoCA9PoSLg W3DlM08YMyNZrh7EJOQLflrc

Question 5:

Find the ratio in which the line segment joining A (1, − 5) and B (− 4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.

Answer:

Let the ratio in which the line segment joining A (1, −5) and B (−4, 5) is divided by x-axis bek : 1

Therefore, the coordinates of the point of division is

We know that y-coordinate of any point on x-axis is 0.
GfqKAVT uldyWJYX1zyaLSO96Sm7ncu6OxYYwzq8weMgZoZjnmTWWzij901 RK620PWgi7uaistJk4bgn7t1aEIsz60xxG3lrPUZ65zwhCmGy9eK e mH2QXERD OWicWjTOW0

Therefore, x-axis divides it in the ratio 1:1.

Division point

Question 6:

If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Answer:
slTR3HZucV4F3g6Aau9tpgRxoHzJ3FG L19R0AiPobxVFUfeja3Dky WI9BXPBDxKUcIfhqMK9ZLUhs IXApUq 8dC8bK z99PCjYFSKtM2mwIpTyeTwGaa1SXXzI6M0inPmFTI

Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a parallelogram ABCD. Intersection point O of diagonal AC and BD also divides these diagonals.

Therefore, O is the mid-point of AC and BD.

If O is the mid-point of AC, then the coordinates of O are
5qc vUG9tt7li3nr zEDVi CFrSTSLNF xDXX 8oyqWAb05m7tv6yO4OtyHQu8Kyx7UKktdb P cpuSrOmoMzS tZWuK0Ba1xRH35IsznpUT6sRevllRL7bS EIWqKd6HtgSyo0

If O is the mid-point of BD, then the coordinates of O are
fCVXszm0 3F8vgMkyIi57SqTp6Mn WiA7OpLsKiMuzjmBRQM

Since both the coordinates are of the same point O,
CydMq8qDkRVZx5YGDjBG3FkVDwFYIrYpt5kTDw6 6HbXMrd5HrXp228hwYnixK

*Question 7:

Find the coordinates of a point A, where AB is the diameter of circle whose centre is  (2, −3) and B is (1, 4)

Answer:

Let the coordinates of point A be (x, y).

Mid-point of AB is (2, −3), which is the center of the circle.

*Question 8:

If A and B are (− 2, − 2) and (2, − 4), respectively, find the coordinates of P such that and P lies on the line segment AB.

Answer:

LvsO20QKaGiWD6C3A WqrGu2glCWzebIzTIXRWF1eWauYWd3zcTcrX2t9qARixFhg NoXX3kLq

The coordinates of point A and B are (−2, −2) and (2, −4) respectively.

Since

Therefore, AP: PB = 3:4

Point P divides the line segment AB in the ratio 3:4.g8 YHlAN99jusA8HvrrqclIo5iNmwI djTbKOSg8b b8GuT Nx7O4EHc6JOHMKQirKpArAybcKLwwqwbM0x01tuL 8B2CADPUCZkkK9 bTT0ibHDAZSP5f gmrrcRMrjO A9Nh4

*Question 9:

Find the coordinates of the points which divide the line segment joining A (− 2, 2) and B (2, 8) into four equal parts.

Answer:z1SFi3mPOvBgNXi8N1cu7zi2k8XjhvMvTpCd4ym7 VhHVF9tuhXFPJX4nsr4p2IFm O QPlyrTSkIvtw1mZFwZaN0fP5Kzz7jmwrXEIiENnQFtkzD6IKbcOKUIS 0106GaLezSM

From the figure, it can be observed that points P, Q, R are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.
zMwdwu9SWKsb1wfJ8H5fkmMJtubsYdYrxRkC8ZI38kJUPvmXccP7ouHrFaWA614pL1Gh an0surgwRE93 fuGPf

*Question 10:

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (− 1, 4) and (− 2, −1) taken in order. [Hint: Area of a rhombus (product of its diagonals)]

Answer:

Let (3, 0), (4, 5), (−1, 4) and (−2, −1) are the vertices A, B, C, D of a rhombus ABCD.



Exercise 7.3

Question 1: 

Find the area of the triangle whose vertices are:

(i)     (2, 3), (− 1, 0), (2, − 4) 

(ii)     (− 5, −1), (3, − 5), (5, 2)

Answer:

(i)     Area of a triangle is given by

fHtiDCY7W4aUe4xcQPxHfokkzlOTu1k1rC9Y553lOrzj9KojKEykgU1cO38 jTSc5zBcErAoc

2Z1IvUXFmzhejFAsOxbI XmZDxU6lWCIJPtlqvv4It7IaIRTxYHtoGrQWGXKL9DLkRIJI Lac7FShuodnX7ny3 KF2Eedb4Xp2gnvZOpN XkpnaaCNRuijNoDDpWKyNw1KmLCCk

(ii)z2nb8chDbTi4HNLcQ1kdOU p5tCUF1emzrpnI4wEEuYyhcz 8qJ1XEGgtgccK qU0rsM1YyEAc2Y2Da0L5Sg3SvqDSexm CHJ38BdZGLaHr4B0gy6XeS0lX2XLE3yyZT308b6jc

Question 2:

Question 2:

In each of the following find the value of ‘k’, for which the points     are collinear. 

(i)    (7, − 2), (5, 1), (3, − k) 

(ii)     (8, 1), (k, − 4), (2, − 5)

Answer:

  1.         For collinear points, the area of the triangle formed by them is zero.

    Therefore, for points (7, −2) (5, 1), and (3, k), area = 0
pRLyXth4Dr1diiJm00OphQQWVjayqpu MDxQ6u50WORWVgmFuX7Ml8ihsRq9uIHljAZg03gNl6OlmTaCqfobsxFoJTCF10kf

  1.         For collinear points, area of triangle formed by them is zero.

    Therefore, for points (8, 1), (k, −4), and (2, −5), area = 0

Question 3:

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, − 1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Answer:
ccafmg WvJrfWJEGkNC2jnuiFwFFDUDBZ2yegD0A6XkhgD9gCqnXLh3 X wd6CiiIsHja8TZPxcj3nsHG3vSBRSXxDbZyAaKwEwRaezplNF0eo9LmxYdyUhFRmwsAt3ETF2 rWg

Let the vertices of the triangle be A (0, −1), B (2, 1), C (0, 3).

Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by

V0izgFHE28fPEKiKZSeEEO23H3PwNl9eQxfpvTYH74ulUjocp6oWwL zfAv5gI3ttBewZxtLqSeLNltWsH Btd AcXLT3 NSgLx8dAFCv

*Question 4:

Find the area of the quadrilateral whose vertices, taken in order, are (− 4, − 2), (− 3, − 5), (3, − 2) and (2, 3)

Answer:

Let the vertices of the quadrilateral be A (−4, −2), B (−3, −5), C (3, −2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.

6yxTBoz2u0r4nw8rkv1ipcVr9VQDhYdwO9 xniS5g5h3jmmK4PwDDSejjCCdHLofl PDkHzo8tkqO53MQ9dRRLiEy921UKBhTjy2esf6beCH4oreQtC

*Question 5:

You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, − 6), B (3, − 2) and C (5, 2)

Answer:

Let the vertices of the triangle be A (4, −6), B (3, −2), and C (5, 2).

Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.

Jw0CbyfpbUOsCpxufo PEjWFb5qwFVWUaUmYRCd231smamGGS8nf8u8DaoLOQpZRsDoW556n yuACN2LJe7FDwZyhnVVT05MwCEYiiaDKLMajsNHlOUoC5NSoFO30JqH8p8YdfI

However, area cannot be negative. Therefore, area of ΔABD is 3 square units.tuFARbHEiLu5ELd zjZ5b1m0DMYHDgQQCIgi8aUXPkF kxf yz3R85K13WIzSuymQ1gg0uv2WJ3haHR R7QqXHiloY81653Y7ZsvkEKbIdZtNV8wV6KB7 ryRr8UlInVEeb bBE

However, area cannot be negative. Therefore, area of ΔADC is 3 square units.

Clearly, median AD has divided ΔABC in two triangles of equal areas.

Exercise 7.4

*Question 1:

Determine the ratio in which the line 2x + y − 4 = 0 divides the line segment joining the points A(2, − 2) and B(3, 7)

Answer:

Let the given line divide the line segment joining the points A(2, −2) and B(3, 7) in a ratio k : 1.

Coordinates of the point of division

This point also lies on 2x + y − 4 = 0

Therefore, the ratio in which the line 2x + y − 4 = 0 divides the line segment joining the points A(2, −2) and B(3, 7) is 2:9.FxM1bY OPMrRluiJ7MuzoyUswBzK Pkt7

*Question 2:

Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

Answer:

If the given points are collinear, then the area of triangle formed by these points will be 0.

1TgK6RMh5CsLEyudT5TDjiEskVt2o4t6J3kOy8vFydmCImc6dYS6LGpezVPngA1UcaL124 wqFIWoNF68scWx7g5KH17dn AY YXjG014C2mmH56euIP5QUTnAP2O3nO kI8qAA

This is the required relation between x and y.

*Question 3:

Find the centre of a circle passing through the points (6, − 6), (3, − 7) and (3, 3).

Answer:

Let O (x, y) be the centre of the circle. And let the points (6, −6), (3, −7), and (3, 3) be representing the points A, B, and C on the circumference of the circle.

4hgH64 x0AX1GQ 1AVCRO8DBKwDm34N j3Eqm83vC08 OB2nQ5wnatKi70rfOAh7jvquRtiDLpq 0vynGLI lkk64t0FvfzzYzfOkDdbDkCq3FlN0ORGZZGkcDgZ6psvvSIwgvg
7mXguAb5lwemZhJ05jDRQD2wORQwmQ0tqLL5P9ZgcIvBRWvweeDspOJAmLhz2xrGvNOiHiv26VBBKomh6ch6NF307D8h S7et plnsQnrCdTxIj2cnDf 6A90b42vJN6jwvoABE

On adding equation (1) and (2), we obtain

cTgxTv4QO Mr3OPczSRdHAc9CNcQh0Ncm8G XRg9wX7E0lHWDiwvAlnuRTxrcVZFxmtXeQ8kZzx0 OAhjR hCGms5aCrqXvOssQrH1nUrQ BzbK9WBckP6o3WmcTQLhdorQQaUU
e5H1cQ2OR9Z8IXPiKb

From equation (1), we obtain

522fM8DxF5l6K3mfzi dxpIHvm0qeCx1iJZZUCpBDbAIlNnR69Wz46Jm 35xe0pGzlufx1RkpCteF71YwAp4o4tjbgEJWz2U Rb5FdS0qreZBGezl66IzOYlqZe1k7O3k8dy zU
inPF9SimMBs9 Q9KAGXVotLeQVK25oMUvFCjgRcoKwdIAGny hLspG1AKy gFrHNC21bLP32AUM3v3sIkl0wzvIE35RILTqDaezS2CzUbfY3i5C2jBTeBDDSTVPSVOgPI NEwRQ
tdOglYF04TIiMWOtl2I0EQBjnpUThuQwod6NARy4x8OsL EusjY7PlyvPzzqhQLKsJ 1GAS3Moa6CN OqEG1KH4Wuvpn02XTL7gifjWLwIvHoD svfTNBRJfsdaisWhv6D6Misw

Therefore, the centre of the circle is (3, −2).

*Question 4:

The two opposite vertices of a square are (− 1, 2) and (3, 2). Find the coordinates of the other two vertices.

Answer:

Let ABCD be a square having (−1, 2) and (3, 2) as vertices A and C respectively. Let (x, y), (x1, y1) be the coordinate of vertex B and D respectively.

We know that the sides of a square are equal to each other.

∴ AB = BC

rvn 3liB ZeTiXfQFEY5dB25Tg5iW7 Y6 rnG5u8ue9takFucRBO6cOFsPw dw 1iSUKXzxiewqR F xveyxTsK6M0oEBn gF5xjJY12mPq 2AXz 7VeI 1h uziOL19h0ERfAYyW7zVYaPOQu1x3B1lz5f6kLN2EKoc9gJOx LgHMkkgsSyIpTOnFiPsjXrh96S0y4cV5BQ5AkAYQ5Qe3K3zCh2PB

We know that in a square, all interior angles are of 90°.

In ΔABC,

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dWCF uRHatMwZcTiXtLNPrirBTCvenlWwDWFl 0emCnvJaBmJWnoj0cNvKUmjbKgZTVqYT 9zxXdL
mQUyax XPyFHrRwiVQZ5slWmoOdkIapRe4F CRW31Fi23mSP1ToRTa0a
USHwyKVKVkU0LRsWmT5RFHCb43i1i7AnbXaXpoTlZ6u9sx3J5D2HW TM08 RE2utu7MsxiHfGXzBJBNclsp 7bN4b5MvI HuHKm7BjsCcunoMcN2MozENLnHqSZr0KKj3SMwBlM
3R0fiYeM6meG82OA3kT241XMRW3GFpTvFhBAqWktcICxeKwEwj6KfX2
fCV6kb0Xu0QXZPi69bKizKYduzcLVF2qrBODxz6jT4aJUYP0jkHN2SwLWDdjNVQBMWwO8DJ AwQiTy4ey6cvvMTB2FDYnnqnNMPgiEe3UbE2P7eauJxrkZ6DFCqHFeUUu5q Uhk

or 4       

We know that in a square, the diagonals are of equal length and bisect each other at 90°. Let O be the mid-point of AC. Therefore, it will also be the mid-point of BD.
F6tm gil07eubeJRzTnseJUq1MlI70Ol1FzVRUfOmlf4aoOAXyli4sxIqN19Y46lv8N

9HehztZwjL 0

If

CdALqD2kLjTI7OONWvfU12n462S lT2ULHKZpvg WXEvjXysmSRXadTRJlh4ebp9Qq57TWE0MhHGpd67T7fj XoNWu64E5AB 3cqfcugf qJcdWfs2TDU18odm9SkYuDNdjTtVo

If

ZAUrzTf4zcoQDX30NLSRn0QJA6aJfcJIuRHIJsfeKka16IpzOMxHmRm XXHclF4GLmK ov ik zyKHwwSnxdQw59fv

Therefore, the required coordinates are (1, 0) and (1, 4).

*Question 5:

The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the following figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
dcJKGbJSAzV7NwD j1NTgTGpNK35HQdI4 Zb8iam1aXxZbZnXw 5vauor5PTiay lXMOpF2Xs7gBt6QTdoKo0B0N6VuM52uKOBxYyNFp5j6N ljkjGSYdLLTQ4SA3mvPLOr22w

  1.  Taking A as origin, find the coordinates of the vertices of the triangle.
  2.     What will be the coordinates of the vertices of Δ PQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe?

Answer:

(i)        Taking A as origin, we will take AD as x-axis and AB as y-axis. It can be             observed that the coordinates of point P, Q, and R are (4, 6), (3, 2), and (6,         5) respectively.

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(ii)        Taking C as origin, CB as x-axis, and CD as y-axis, the coordinates of vertices         P, Q, and R are (12, 2), (13, 6), and (10, 3) respectively.

1IbiFIk763a7ix5UWjywWS1wwLJd5afmq98b22rQ0PcjDdOHNF jnKbUQLdSfmpVi1pwu1UFAw6MC8bTt1RiNuAu6uGbN1dZXGBRH4aqDPNN0IsFwY6uVkS VsIrlBRyAY9H4lk

It can be observed that the area of the triangle is same in both the cases.

*Question 6:

The vertices of a ΔABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect

sides AB and AC at D and E respectively, such that Calculate the area of the ΔADE and compare it with the area of ΔABC. (Recall Converse of basic

proportionality theorem and Theorem 6.6 related to ratio of areas of two similar triangles)

Answer:
nvIiMJXetd56HW0wj6g41OEYo1aNJXIgXlyA1zsGdN1J4nrsPD hio8WaVt4wIgRDaCsobbro9ckc2YEJFQOzfRFB2C4Esblui2Yt0FduKt9vAuTMNkuUgJse6shBpYpm0lm5oY

Given that,wc2AlMvFKhtAT jQv

8E2DSzTwHNqruIzwBaGymnauLPN4qADL BM6AdsU3XOeVoxlPZxF7

Therefore, D and E are two points on side AB and AC respectively such that they divide side AB and AC in a ratio of 1:3.
STtdAaqUctMPfWJbzfJDhPr4Fi45dzIdYrhLY4WRFg2VitojQ571sMRFVnWB Ja7zETkYyKfdacJ5gN9nM4YQdMN7FUJ3ndQ5QzNUPk M tyRAanLddibYuX7rzvytkMToyBLd0ogs96flCgAIkY7LNCI6 cie3ZYuaNfGmFvBWe3Ixqk17H68TNwvDw uoMhX4pApPiPlnttqWZc 3 t8xMOiaMvCp HIEwLQO4 iPv5ucyLniJNllXvWTPcJVRlsvJbJ3HiAbFeKzDJIQA 6Agxx fVq9CuTnfo4xwm964D3VVSJ2txKr3gyt8MIQCFY dEXPnn7nYmc3gJcJy rbtsXVbt7h DK6QV7tHREHIQUgLPs8 zri0 wpt2iYFULI8

fQx3p78 U aFo8o1W6TlQPP1X8vI3HT8pyVJQMcwpCYBCi h2q7w8WMda8ZDmRQl3CAdWx0anlkt5FuT7VHU6vgQnbi

Clearly, the ratio between the areas of ΔADE and ΔABC is 1:16.

Alternatively,

We know that if a line segment in a triangle divides its two sides in the same ratio, then the line segment is parallel to the third side of the triangle. These two triangles so formed (here ΔADE and ΔABC) will be similar to each other.

Hence, the ratio between the areas of these two triangles will be the square of the ratio between the sides of these two triangles.

Therefore, ratio between the areas of ΔADE and ΔABC

*Question 7:

    Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ΔABC.

  1.         The median from A meets BC at D. Find the coordinates of point D.
  1.         Find the coordinates of the point P on AD such that AP: PD = 2:1
  1.     Find the coordinates of point Q and R on medians BE and CF respectively         such that BQ: QE = 2:1 and CR: RF = 2:1.
  2.     What do you observe?
  1.         If A(x1, y1), B(x2, y2), and C(x3, y3) are the vertices of ΔABC, find the coordinates         of the centroid of the triangle.

Answer:

  1.         Median AD of the triangle will divide the side BC in two equal parts.

    Therefore, D is the mid-point of side BC.v7a3plWGdlHauCXPIQfBAIppk EiALz tl8KYwTtKauPNle2zBcoovFJVQX30gRSr

  1.         Point P divides the side AD in a ratio 2:1.pOv4 WtAawt7q7N4uYHdhz5XoKBf5R1TXSL lNXxNFoHIQy4Vo997O 9ID EdnD63UAOTZrT5fQGykUH cmGaelb916NcWca2Au8YPe7Mqhl4K0ZSLvLjylI2eI6b rd7 2tzZs
  1.     Median BE of the triangle will divide the side AC in two equal parts.

    Therefore, E is the midpoint of side AC.

    Point Q divides the side BE in a ratio 2:1.OLIuB88WOaVNeBR4cGXVJ qnnk1 VIqOBPmhaxv2bwU7KxB7pfl8y 15fMmJ0RGNYXKC2wK9voKYh6Pe Xj1ZnoAbeXLNlR5cfH70y5S RCmeOfULKAFBGOuzgfjLPsK4N6dLL4

Median CF of the triangle will divide the side AB in two equal parts. Therefore, F is the mid-point of side AB.
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Point R divides the side CF in a ratio 2:1.

  1.     It can be observed that the coordinates of point P, Q, R are the same.6tKnbF5AFFwPF t2hDKEJP3zFTPxRDjxntP7r2q wHL91kRx465dh33GSKUtMhH2Ljf4ljow4sn9ve5zQ7VhbZvCPu4lN4j4mt8q00Ztq27A42 8as55sVPmHpEz7gUTPGm3SaQ

    Therefore, all these are representing the same point on the plane i.e., the     centroid of the triangle.

  1.         Consider a triangle, ΔABC, having its vertices as A(x1, y1), B(x2, y2), and             C(x3,y3).

    Median AD of the triangle will divide the side BC in two equal parts. Therefore,     D is the mid-point of side BC.

    Let the centroid of this triangle be O.

    Point O divides the side AD in a ratio 2:1.sY3CyZu2YX 8IYLGYCFboPkT 9eltaA5fB8kSxHkVEwVoZa9UpAO45mnOFq2Czaq3l6VfHOTBILCUE drhbTwXj6j4Woo5mw3HpqLHrTJkxC5Su5jDHzhZ4diToBpHK m1v8y6o

*Question 8:

ABCD is a rectangle formed by the points A (− 1, − 1), B (− 1, 4), C (5, 4) and D (5, − 1). P, Q, R and S are the mid-points of AB, BC, CD, and DA respectively. Is the quadrilateral PQRS is a square? a rectangle? or a rhombus? Justify your answer.

Answer:

HNa

It can be observed that all sides of the given quadrilateral are of the same measure. However, the diagonals are of different lengths. Therefore, PQRS is a rhombus.

This chapter comes under unit 6 and has a weightage of 6 marks in the board examination. There will be one mark MCQ question, 2mark reasoning questions, and 3 marks short answer questions. This chapter has fundamental concepts that lay the foundation for your future studies.

Sub-topics of class 10 chapter 7 Coordinate Geometry

  • Introduction to Coordinate Geometry
  • Distance Formula
  • Section Formula
  • Area of the Triangle

List of Exercise from Class 10 Maths Chapter 7 Coordinate Geometry

Exercise 7.1– 10 Questions which includes 8 practical based questions, 2 reasoning questions
Exercise 7.2– 10 Questions which includes 8 long answer questions, 2 short answer questions
Exercise 7.3– 5 Questions which includes 3 long answer questions, 2 practical based questions
Exercise 7.4– 8 Questions which includes 6 long answer questions, 1 practical based question, 1 reasoning question   NCERT Solutions are carefully drafted to assist the student in scoring good marks in the examination. It provides you with much-needed problem-solving practise.

This chapter deals with finding the area between two points whose coordinate values are provided. For instance the area of a triangle. This chapter has some basic concepts like the area of a triangle, rhombus, the distance between sides, and intersections. This chapter teaches you the relationship between numerical and geometry and their application in our daily lives.

These NCERT Solutions for Class 10 Maths have different types of questions and their answers which will help you with alternate solutions, diagrammatic representation. The solutions provided here are written in simple language with apt information. By studying these NCERT solutions thoroughly students will be able to solve complex problems easily.

Key Features of NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

  • The solution has all the exercise questions provided in the NCERT textbook.
  • Diagrammatic representations and alternate methods will help you understand the concepts thoroughly.
  • Solving these NCERT Solutions will make you acquainted with important formulas and standards.
  • This NCERT Solution has different examples that will help you relate to real-life examples to relate geometry and numerical.

Conclusion

The NCERT Solutions for Class 10th Maths Chapter 7 – Coordinate Geometry, provided by Swastik Classes, offer a comprehensive understanding of the fundamental concepts of coordinate geometry. The chapter covers various topics such as plotting points on a graph, finding the distance between two points, and dividing a line segment into two parts.
The solutions provided by Swastik Classes are well-structured and easy to understand, with step-by-step explanations for all the exercises. The solutions are designed to help students develop their problem-solving skills and gain a deeper understanding of the concepts covered in the chapter.
Coordinate geometry is an essential part of the mathematics curriculum and is widely used in various fields such as physics, engineering, and computer science. The solutions provided by Swastik Classes will help students build a strong foundation in coordinate geometry and prepare them for more advanced mathematical concepts.
Overall, the NCERT Solutions for Class 10th Maths Chapter 7 – Coordinate Geometry, provided by Swastik Classes, are an excellent resource for students who want to excel in the subject. The solutions will not only help them score well in their exams but also prepare them for more advanced topics in mathematics. It is crucial to have a clear understanding of the concepts covered in this chapter as it forms the foundation for more advanced concepts in coordinate geometry.

How NCERT Solutions for Class 10 Maths Chapter 7 helpful for board exams?

NCERT Solutions for Class 10 Maths Chapter 7 provides answers with detailed descriptions as per term limit particularised by the Board for self-evaluation. Solving these solutions will provide excellent practice for the students so they can finish the paper on time. So, it’s clear that the NCERT Solutions for Class 10 Maths Chapter 7 are essential to score high in examinations. Students can get acquainted with writing exams and will be able to face exams more confidently.

Mention the topics that are covered in NCERT Solutions for Class 10 Maths Chapter 7?

The topics that are covered in NCERT Solutions for Class 10 Maths Chapter 7 are introduction to coordinate geometry, distance Formula, section formula and area of the triangle. By learning these NCERT solutions thoroughly students will be able to solve complex problems easily.

Is it necessary to practice all the exercises present in Chapter 7 of NCERT Solutions for Class 10 Maths?

Yes, it is compulsory to practice all the exercises present in Chapter 7 of NCERT Solutions for Class 10 Maths. Because all exercises contain numerous questions to solve which may come in their finals. This makes them more confident towards the syllabus they have.


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