# NCERT Solutions for Class 10 Maths Chapter 7

NCERT solutions for class 10 Maths chapter 7 exercise 7.4, 7.3, 7.2 & 7.1 (प्रश्नावली 7.1, 7.2, 7.3 & 7.4) of coordinate geometry for UP Board Schools 2019-20 as well as CBSE, MP Board Schools in Hindi Medium and English medium PDF format to free download. Download NCERT Solutions Apps based on updated NCERT Solutions for the new session 2019-2020. There is overall summery about coordinate geometry for class 10, which will help the students to know more about this chapter.

 Class 10: Maths – गणित Chapter 7: Coordinate Geometry ## NCERT solutions for class 10 Maths chapter 7

Go Back to Class 10 Maths Main Page

### Class 10 Maths Solutions – Coordinate Geometry

#### English Medium & Hindi Medium Solutions

• 10 Maths Chapter 7 Exercise 7.1 Solutions
• 10 Maths Chapter 7 Exercise 7.2 Solutions
• 10 Maths Chapter 7 Exercise 7.3 Solutions
• 10 Maths Chapter 7 Exercise 7.4 Solutions
• NCERT Book in Hindi & English Medium

#### 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]

In class 9, we have studies 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.

#### 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.

#### 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.
• The distance of the line segment joining two points (x1, y1) and (x2, y2) is given by: • The co-ordinates of a point, which divides the line segment joining two points (x1, y1) and (x2, y2) in a ratio m : n internally are given by: • The co-ordinates of the mid-point of the line segment joining two points (x1, y1) and (x2, y2) are given by: • The co-ordinates of the centroid of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) are given by: • If ABC be any triangle whose vertices are A(x1, y1), B(x2, y2) and C(x3, y3), then area of triangle is given by: 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.

#### उस त्रिभुज का क्षेत्रफल ज्ञात कीजिए जिसके शीर्ष हैं (2, 3) ,(-1, 0) ,(2, -4).

त्रिभुज के शीर्ष A(2, 3),B(-1, 0) और C(2,-4) हैं।
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

#### Find the value of ‘k’, for which the points (7, -2), (5, 1) and (3, k) are collinear.

A(7, -2), B(5, 1), C(3, k)
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

#### y का वह मान ज्ञात कीजिए, जिसके लिए बिंदु P(2, -3) और Q(10, y) के बीच की दूरी 10 मात्रक है।

बिंदु P(2, -3) और Q(10, y) के बीच की दूरी 10 मात्रक है।
⇒ √((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