# NCERT Solutions for Class 10 Maths Chapter 7

NCERT solutions for class 10 Maths chapter 7 coordinate geometry all exercises (Ex. 7.1, 7.2, 7.3, 7.4) for all Boards based on NCERT Books. These solutions are updated for academic session 2020-2021 for UP Board as well as CBSE, MP Board Schools in Hindi Medium and English medium in PDF format to free download.

UP Board High School students are now using NCERT Books for school exams. Download NCERT Solutions Apps 2020-21 based on updated NCERT Solutions for the new session 2020-2021. There is overall summery about coordinate geometry for class 10, which will help the students to know more about this chapter.## NCERT solutions for class 10 Maths chapter 7

Class: | 10 |

Subject: | Maths – गणित |

Chapter 7: | Coordinate Geometry |

### 10th Maths Chapter 7 Solutions

NCERT solutions for class 10 Maths chapter 7 coordinate geometry exercises from 7.1 to 7.4 are given below to free download. All the contents are in updated form for academic session 2020-21. Offline Apps based on these solutions are also for new session. Join the Discussion Forum to discuss your doubts and respond the questions asked by your friends.

### 10th Maths Chapter 7 Exercise 7.1

### 10th Maths Chapter 7 Exercise 7.2

### 10th Maths Chapter 7 Exercise 7.3

### 10th Maths Chapter 7 Exercise 7.4

#### Class 10 Maths Exercise 7.1 & 7.2 Solutions in Video

#### Class 10 Maths Exercise 7.3 & 7.4 Solutions in Video

### NCERT Solutions for Class 10 Maths Chapter 7

- NCERT Solutions 10th Maths
- NCERT Solutions 10th Maths
- NCERT Solutions 10th Maths
- NCERT Solutions 10th Maths

##### What are the Objectives of Coordinate Geometry in class 10 Maths?

Objectives of Coordinate Geometry

To find the distance between two different points whose co-ordinates are given and finding the co-ordinates of a point, which divides the line segment joining two points in a given ratio internally. To find the co-ordinates of the mid-point of the join of two points to get the co-ordinates of the centroid of a triangle with given vertices.

##### Give some Important Results in Coordinate Geometry?

Important Results in Coordinate Geometry

The co-ordinates of the origin are (0, 0)

The y co-ordinate of every point on the x-axis is 0 and the x co-ordinate of every point on the y-axis is 0.

The two axes XOX’ and YOY’ divide the plane into four parts called quadrants.

#### Previous Years Questions

ONE MARK QUESTIONS

1. If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k? [CBSE 2017]

THREE MARKS QUESTIONS

1. The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is (7/2, y), find the value of y. [CBSE 2017]

2. Show that triangle ABC, where A(-2, 0), B(2, 0), C(0, 2) and triangle PQR where (-4, 0), Q(4, 0), R(0, 4) are similar triangles. [CBSE 2017]

FOUR MARKS QUESTIONS

1. If a≠b≠0, prove that the points (a, a^2), (b, b^2), (0, 0) will not be collinear. [CBSE 2017]

##### Know about Coordinate Geometry

In Coordinate geometry, we study that the distance of a point from the y-axis is called its x-coordinate, or abscissa (abscissa is a Latin word which means cut off) and the distance of a point from the x-axis is called its y-coordinate, or ordinate (ordinate is a Latin word which means keep it in order). Abscissa and ordinate collectively forms coordinate of a point in Cartesian system. The coordinates of a point on the x-axis are of the form (x, 0), and of a point on the y-axis are of the form (0, y). For more question on coordinate geometry, go through NCERT exemplar problems for Class 10 Maths.

###### Historical Facts!

Rene Descartes (1596 – 1650) was a French philosopher, mathematician whose work ‘La geometrie’ includes his application of algebra to geometry from which we now have Cartesian geometry.

### Important Questions on Class 10 Maths Chapter 7

त्रिभुज के क्षेत्रफल व्यंजक

= 1/2 [x_1 (y_2 – y_3 ) + x_2 (y_3 – y_1 ) + x_3 (y_1 – y_2 )] द्वारा

त्रिभुज ABC का क्षेत्रफल

= 1/2 [ 2 { 0 – (- 4)} + (-1) {(-4) – 3} + 2(3 – 0)]

= 1/2 [8 + 7 + 6]

= 21/2

= 10.5 वर्ग मात्रक

Area of triangle formed by three collinear points is zero.

Therefore, the area of triangle ABC = 0

⇒ 1/2 [7(1 – k) + 5{k – (-2)} + 3(-2 -1)] = 0

⇒ 7 – 7k + 5k + 10 – 9 = 0

⇒ -2k = -8

⇒ k = 4

⇒ √((10 – 2)^2 +[y – (-3)]^2 ) = 10

⇒ √(64 + y^2 + 9 + 6y) = 10

दोनों ओर वर्ग करने पर

64 + y^2 + 9 + 6y = 100

⇒ y^2 + 6y – 27 = 0

⇒ y^2 + 9y – 3y – 27 =0

⇒ y(y + 9) – 3(y + 9) = 0

⇒ (y + 9)(y – 3) = 0

⇒ (y + 9) = 0 या (y – 3) = 0

⇒ y = – 9 या y = 3