NCERT Solutions for Class 9 Maths Chapter 3: Coordinate Geometry

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The NCERT Solutions for class 9 Maths Chapter 3 Coordinate Geometry have been provided in the article below. Coordinate Geometry deals with geometrical problems solved using coordinate systems and points.

Class 9 Maths Chapter 3 Coodinate Geometry belongs to Unit 3 Coordinate Geometry having a weightage of 04 in the Class 9 Maths Examination. NCERT Solutions for Class 9 Maths for Chapter 3 cover the following important concepts: 

Download: NCERT Solutions for Class 9 Maths Chapter 3 pdf


NCERT Solutions for Class 9 Maths Chapter 3

The Chapter 3 Class 9 Maths are given below:

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Important Topics in Class 9 Maths Chapter 3 Coordinate Geometry

Important Topics in Class 9 Maths Chapter 3 Coordinate Geometry are elaborated below: 

  • Distance Formula of Coordinate Geometry

Distance formula of coordinate geometry, derived from the Pythagoras Theorem, is used to determime the distance between any 2 given points. These points are usually located on an x-y coordinate plane. Distance Formula of coordinate geometry can be written as, AB = √[(x2-x1)²+(y2-y1)²]

Example: Consider coordinate A (-4,0) and B (0,3). With the given coordinates, determine the distance between these two points.

Solution: Considering the coordinates ofA = (-4,0) = (x1, y1)
And the coordinates of B = (0,3) = (x2,y2)
Now by using Distance Formula, we can get,
AB = √{(x2-x1)²+(y2-y1)²} = √{[0-(-4)]²+ (3-0)²}
= √(4²+3²)} = √(16+9) = √25
= 5 units

  • Section Formula of Coordinate Geometry

Section Formula helps to determine the coordinates of a point that assists division of the line joining two points in a ratio. The equation occurs either internally or externally.

Section Formula of Coordinate Geometry can be classified into two parts:

  1. Internal Section Formula: \(P(x,y) = (\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n})\)
  2. External Section Formula: \(P(x,y) = (\frac{mx_2-nx_1}{m-n} , \frac{my_2-ny_1}{m-n})\)
  • Ordinate 

Ordinate represents the coordinate values present on the y-axis on a coordinate system. It is the second component of an ordered pair.

Example: What will be the abscissa and ordinate of a point with coordinates (8,12)?

Solution: The coordinates of (8, 12) are:
Abscissa: 12
And, Ordinate: 8

  • Quadrant

Quadrants are the parts that form when two coordinate axes of a plane tend to intersect with one another at an angle of 90 degree. The intersection these two lines experience is called a point of reference. 

Example: Highlight the quadrants where coordinate points (-2,7) take place? 

Solution: (-2,7) falls in the second quadrant. Here, the value of the x axis becomes negative. 

  • Cartesian System

Cartesian System, derived from the number line, is used to label points in a plane. The cartesian form is derived from the number line. It has two perpendicular lines named as X-axis and Y-axis.

Important points to remember in Cartesian System:

  • The point of intersection of both the axes is called the origin. It has coordinates (0, 0).
  • An infinite number of points can be plotted on a cartesian coordinate plane.
  • Points that fall on any of the number lines don’t belong to any quadrant.

NCERT Solutions for Class 9 Maths Chapter 3 Exercises:

The detailed solutions for all the NCERT Solutions for Coordinate Geometry under different exercises are:

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CBSE X Related Questions

  • 1.
    Find the smallest value of $p$ for which the quadratic equation $x^2 - 2(p + 1)x + p^2 = 0$ has real roots. Hence, find the roots of the equation so obtained.


      • 2.

        In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD.


          • 3.
            Using prime factorisation, find the HCF of 144, 180 and 192.


              • 4.
                In a trapezium \(ABCD\), \(AB \parallel DC\) and its diagonals intersect at \(O\). Prove that \[ \frac{OA}{OC} = \frac{OB}{OD} \]


                  • 5.
                    In \(\triangle ABC, DE || BC\). If AE = (2x + 1) cm, EC = 4 cm, AD = (x + 1) cm and DB = 3 cm, then value of x is
                     △ABC,DE||BC. If AE = (2x + 1) cm, EC = 4 cm, AD = (x + 1)

                      • 1
                      • \(\frac{1}{2}\)
                      • --1
                      • \(\frac{1}{3}\)

                    • 6.

                      There is a circular park of diameter 65 m as shown in the following figure, where AB is a diameter. An entry gate is to be constructed at a point P on the boundary of the park such that distance of P from A is 35 m more than the distance of P from B. Find distance of point P from A and B respectively.

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