# Coordinate Geometry: Class 10 Mathematics NCERT Chapter 7

**Key Features of NCERT Material for Class 10 Mathematics Chapter 7 – Coordinate Geometry **

In the last chapter 6, you learned about everything about triangles. In this chapter, you will learn about Coordinate Geometry.

Position of a point P in the Cartesian plane as for coordinate axes is shown by the arranged pair (x, y).

The line X’OX is said to be the X-axis and YOY’ is known as the Y-axis. The portion of intersection of the X-axis and Y-axis is known as the root O and the coordinates of O are (0, 0).

The perpendicular distance of a point P from the Y-axis is the ‘x’ coordinate and is known as the abscissa.

The perpendicular distance of a point P from the X-axis is the ‘y’ coordinate and is known as the ordinate.

Indications of abscissa and ordinate in various quadrants are as given in the outline:

**Distance Formula **

Distance between Two differnt Points on the Same Coordinate Axes

The distance between two points which are on a similar axis (x-axis or y-axis), is given by the distinction between their ordinates on the off chance that they are on the y-axis, else by the contrast between their abscissa on the off chance that they are on the x-axis.

Distance AB is equal to 6 – (- 2) = 8 units

Distance CD is equal to 4 – (- 8) = 12 units

Distance between Two Points Using Pythagoras Theorem

Discovering distance between 2 points utilizing

**Pythagoras Theorem **

Let P(x1, y1) and Q(x2, y2) be any two points on the cartesian plane.

Attract lines corresponding to the tomahawks through P and Q to meet at T.

ΔPTQ is a right triangle angled at T.

By Pythagoras Theorem,

PQ2 = PT2 + QT2

= (x2 – x1)2 + (y2 – y1)2

PQ

=√[x2 – x1)2 + (y2 – y1)2]

**Distance Formula **

Distance between any distinct two points (x1, y1) and (x2, y2) is given by

d = √[x2 – x1)2+(y2 – y1)2]

Where d is the distance between the points (x1,y1) and (x2,y2).

Any point on the X-axis is in the form(x, 0).

Any point on the Y-axis is in the form(0, y).

The distance between two distinctive points P(x1, y1) and Q (x2, y2) is given by

PQ =

**Section formula**

. The coordinates of the point which separates the line segment joining the points A(x1, y1) and B(x2, y2) inside in the ratio m : n are:

The above formula is section equation. The proportion m: n can likewise be composed as : 1 or k : 1, The

coordinates of P can likewise be composed as P(x,y) =

The midpoint of the line segment connecting the points P(x1, y1) and Q(x2, y2) is

Here m : n = :1.

Area of a Triangle. The area of a triangle shaped by points A(x1 y1), B(x2, y2) and C(x3, y3) is given by | ∆ |,

where ∆ =

where ∆ speaks to the outright worth.

Three points are collinear if |A| = 0.

On the off chance that P is centroid of a triangle, at that point the middle partitions it in the proportion 2 :1. Coordinates of P are given by

Area of a quadrilateral, ABCD is equal to ar(∆ABC) + ar(∆ADC)

**Discovering ratio given the points **

To discover the ratio wherein a given point P(x, y) separates the line segment joining A(x1, y1) and B(x2, y2),

Accept that the proportion is k : 1

Replace the ratio in the section formula for any of the coordinates to get the estimation of k.

\(x=\frac{kx_{2}+x_{1}}{k+1}\)

When x1, x2 and x are known, k can be determined. The equivalent can be determined from the y-coordinate too.

**MidPoint **

The midpoint of any line segment partitions it in the proportion 1 : 1.

The coordinates of the midpoint(P) of line segment joining A(x1, y1) and B(x2, y2) is given by

\(p(x,\ y)=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2} )\)