NCERT CBSE Class 10th Mathematics Chapter 7: Coordinate Geometry

Safalta Expert Published by: Sylvester Updated Wed, 22 Jun 2022 06:46 PM IST

Highlights

NCERT CBSE Class 10th Mathematics Chapter 7: Coordinate Geometry

The seventh chapter in Mathematics textbook is 'Coordinate Geometry'.
  • Position of a point P in the Cartesian plane with respect to co-ordinate axes is represented by the ordered pair (x, y).
    Coordinate Geometry Class 10 Notes Maths Chapter 7 Q1.1
  • The line X’OX is called the X-axis and YOY’ is called the Y-axis.
  • The part of intersection of the X-axis and Y-axis is called the origin O and the co-ordinates of O are (0, 0).
  • The perpendicular distance of a point P from the Y-axis is the ‘x’ co-ordinate and is called the abscissa.
  • The perpendicular distance of a point P from the X-axis is the ‘y’ co-ordinate and is called the ordinate.
  • Signs of abscissa and ordinate in different quadrants are as given in the diagram:
    Coordinate Geometry Class 10 Notes Maths Chapter 7 Q1.2
  • Any point on the X-axis is of the form (x, 0).
  • Any point on the Y-axis is of the form (0, y).
 

The topics discussed in this chapter are as follows:

  • Points on a Cartesian Plane
  • Distance Formula
  • Section Formula
  • Mid point
  • Points of Trisection
  • Centroid of a Triangle
  • Area from Coordinates

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The midpoint of a line segment joining two points A(2, 4) and B(-2, -4) is

As per midpoint formula, we know;

x-coordinate of the midpoint = [2 + (-2)]/2 = 0/2 = 0

y-coordinate of the midpoint = [4 + (-4)]/2=0/2=0

Hence, (0, 0) is the midpoint of AB.

The distance of point A(2, 4) from the x-axis is

The distance of a point from the x-axis is equal to the ordinate of the point which is 4 units.

The point which divides the line segment of points P(-1, 7) and (4, -3) in the ratio of 2:3 is

By section formula we know:

x = [(2 × 4) + (3 × (-1))]/(2 + 3) = (8 – 3)/5 = 1

y = [(2 × (-3)) + (3 × 7)]/(2 + 3) = (-6 + 21)/5 = 3

Hence, the required point is (1, 3).

The coordinates of a point P, where PQ is the diameter of a circle whose centre is (2, – 3) and Q is (1, 4) is

By midpoint formula, we know;

[(x + 1)/2, (y + 4)/2] = (2, -3) {since, O is the midpoint of PQ}

By equating the corresponding coordinates,

(x + 1)/2 = 2

x + 1 = 4

x = 3

And

(y + 4)/2 = -3

y + 4 = -6

y = -10

So, the coordinates of point P is (3, -10).

The distance of the point P(–6, 8) from the origin is

We know that the distance of a point P(x, y) from the origin is √(x2 + y2).

Thus, the distance of the point P(-6, 8) from the origin = √[(-6)2 + (8)2]

= √(36 + 64)

= √100

= 10 units

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