Coordinate Geometry

Introduction to Coordinate Geometry

Coordinate Geometry is an algebraic tool to study geometry of figures. Coordinate Geometry helps us to study geometry using algebra, and understand algebra with the help of geometry. This is the cause that Coordinate Geometry is widely applied in the various fields, such as physics, engineering, navigation, seismology and art, etc.

What is Coordinates?

A pair of coordinate axes enables us to locate the position of a point on a plane.

The distance of a point from y–axis is called its x–coordinate and the distance of a point from x–axis is called y–coordinate.

x-coordinate is known as abscissa and y–coordinate is known as ordinate also.

Form of coordinates

The coordinates of a point on the x–axis are of the form (`x`, 0).

And the coordinates of a point on the y–axis are of the form (`y`, 0).

Sign of coordinates

(a) The point above the x–axis is considered as positive (+) and below the x–axis is considered negative (–).

(b) The sign right to the y–axis is considered as positive (+) and left to the y–axis is considered as negative (–).

Example of coordinates

In the given graph,

(1) Point O is the origin of plane.

Thus, its coordinates are (0,0).

This means that, distance on x–axis = 0

And, the distance on y–axis = 0

(2) Point A lies on the x–axis.

Thus, its coordinates are (4,0).

This means distance of point on x–axis = 4

And distance of point on y–axis = 0

The coordinates (4,0) means the point lies 4 unit right on x–axis.

(3) Point B is on the y–axis

Thus, the coordinates of point B = (0,7)

This means the distance of point on x–axis = 0

And the distance of point on y–axis = 7

Since, distance on x–axis = 0, thus point B lies on y–axis.

(4) The coordinates of point C are (5,4)

This means the distance of point on x–axis = 5 unit

And distance of point on y–axis = 4 unit

(5) The coordinates of point D are (2,6)

This means the distance of point D on x–axis = 2 unit

And distance of point D on y–axis = 6 unit

Example of coordinates

In the given plane there are four points A, B, C and D are given.

(a) A (–3, 2) means the point A lies 3 units left to the y–axis and 2 units above the x–axis.

(b) B (–2,–1)

This means the point B lies 2 units left to the y–axis and 1 unit below the x–axis.

(c) C (1,2)

This means the point C lies 1 unit right to the y–axis and 2 unit above the x–axis.

(d) D (1–, 2)

This means the point lies 1 unit right to the y–axis and 2 unit below the x–axis.

distance formula

distance formula is used to find out the distance between two points in a plane.

If there are two points P(x₁, y₁) and Q(x₂, y₂), then distance between P and Q,

i.e. PQ `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Solution of NCERT Exercise 7.1

Question (1) Find the distance between the following pairs of points:

(i) (2, 3), (4, 1)

Solution:

Given, two points are (2, 3) and (4, 1)

Thus, distance between given points = ?

Here, x₁ = 2, y₁ = 3 and

x₂ = 4 and y₂ = 1

We know that, distance between two points PQ `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Thus, distance between given points

`=sqrt((4-2)^2+(1-3)^2)`

`=sqrt(2^2+(-2)^2)`

`=sqrt(4+4)`

`= sqrt(8) = sqrt(2xx4)`

`=2sqrt2`

Thus, distance between given points `=2sqrt2` Answer

(ii) (–5, 7), (–1, 3)

Solution:

Here given points are (–5, 7), (–1, 3)

Thus distance between them = ?

Here, we have

x₁ = –5, and y₁ = 7

And x₂ = –1 and y₂ = 3

We know that, distance between two points in a plane, PQ `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Thus, distance between given points `=sqrt((-1-(-5))^2(3-7)^2)`

`=sqrt((-1+5)^2+(-4)^2)`

`=sqrt((4)^2+16)`

`=sqrt(16+16) = sqrt(32)`

`=sqrt(16xx2)`

`=4sqrt2`

Thus, distance between given points `=4sqrt2` Answer

(iii) (a, b), (–a, –b)

Solution :

Given points are (a, b), (–a, –b)

Thus, distance between them = ?

Here, we have, x₁ = a, y₁ = b

And, x₂ = –a, y₂ = –b

We know that, distance between two points in a plane, PQ `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Thus, distance between given points

`=sqrt((-a-a)^2+(-b-b)^2)`

`=sqrt((-2a)^2+(-2b)^2)`

`=sqrt(4a^2+4b^2)`

`=sqrt(4(a^2+b^2))`

`=2sqrt(a^2+b^2)`

Thus, distance between given points `=2sqrt(a^2+b^2)` Answer

Question (2) Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Solution:

(a) Distance between the points (0, 0) and (36, 15)

Given points are (0, 0) and (36, 15)

Thus, distance between these given points = ?

Here, we have

x₁ =0, y₁ =0

And, x₂ = 36, y₂ = 15

Here it is clear that one point is at the origin.

Now, we know that, distance of a point P(x, y) from the origin O (0, 0)

OP`=sqrt(x^2+y^2)`

Thus, distance between given points in question

`=sqrt(36^2+15^2)`

`=sqrt(1296+225)`

`=sqrt(1521)=39`

Thus, distance between given points = 39 unit Answer

(b) the distance between the two towns A and B discussed in Section 7.2

The figure given in section 7.2 (in book) is as follows

From the figure it is clear that, the coordinates of point A = (0, 0) and

the coordinates of point B = (36, 15)

This is the same figure as solved in section (a) above.

Thus, Distance between two town A and B = 39 km Answer

MCQs Test

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