# NCERT Solutions for Class 10 Maths Chapter 7

NCERT solutions for class 10 Maths chapter 7 coordinate geometry all exercises for UP Board Schools 2020-21 as well as CBSE, MP Board Schools in Hindi Medium & English medium PDF format to free download.

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

##### 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