Class 8 – Co-Ordinate Geometry

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Co-Ordinate Geometry
  • Distance Formula and Section Formula
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Introduction: You must have searched for your seat in a cinema hall, a stadium, or a train. For example, seat H-4 means the fourth seat in the Hth row. In other words, H and 4 are the coordinates of your seat. Thus, the geometrical concept of location is represented by numbers and alphabets (an algebraic concept).

Also a road map gives us the location of various houses (again numbered in a particular sequence), roads and parks in a colony, thus representing algebraic concepts by geometrical figures like straight lines, circles and polygons.

The study of that branch of Mathematics which deals with the interrelationship between geometrical and algebraic concepts is called Coordinate geometry or Cartesian geometry in honor of the French mathematician Rene Descartes.

Cartesian plane: It is two dimensional representation of a plane in which a point P is referred with respect to two perpendicular lines intersecting at a point O called the origin.

If  and  be two perpendicular lines intersecting at a point O, then

x-axis: The line  is called x-axis.

y-axis: The line  is called y-axis.

Origin: The point O is called the origin.

Coordinate axes: x-axis and y-axis together are called coordinate axes.


Cartesian coordinates of a point: The ordered pair of perpendicular distance from both axis of a point P lying in the plane is called Cartesian coordinates of P. The Cartesian coordinates of P is written as P(x, y) where x is called abscissa (distance of a point from y-axis) or x-coordinate of P and y is called ordinate (distance of a point from x-axis) or y-coordinate of P.

The above system of coordinating an ordered pair (x,y) with every point in a plane is called the Rectangular Cartesian coordinate system.

It follows from the above discussion that corresponding to every point P in the Euclidean plane there is a unique ordered pair (x,y) of real numbers called its Cartesian coordinates. Conversely, when we are given an ordered pair (x,y) and a Cartesian coordinate system, we can determine a point in the Euclidean plane having its coordinates (x,y). For this we mark off a directed line segment OM = x on the x-axis and another directed line segment ON = y on y-axis. Now draw perpendiculars at M and N to X and Y axes respectively. The point of intersection of these two perpendiculars determines point P in the Euclidean space having coordinates (x,y).

Thus, there is one –to-one correspondence between the set of all ordered pairs (x,y) of real numbers and the points in the Euclidean plane.



Quadrants: The two axes divide the plane into four parts called quadrants.


Sign of x-coordinate

Sign of y-coordinate





(+, +)



(–, +)


(–, –)



(+, –)

The coordinates of the origin are taken as (0,0). The coordinates of any point on x-axis are of the form (x,0) and the coordinates of any point on y-axis are of the form (0,y). Thus, if the abscissa of a point is zero, it would lie somewhere on the y-axis and if its ordinate is zero it would lie on x-axis.

It follows from the above discussion that by simply looking at the coordinates of a point we can tell in which quadrant it would lie.

Example: Plot the points (–3, 0), (2, 3), (–4, 3)and (3, –5) in a rectangular coordinate system.

Solution: As shown in the figure,  and  are rectangular system of coordinate axis.

The axis are labelled with positive numbers along OX and OY, and with negative numbers along  and .

The first point (–3, 0) is represented by A, the second point (2, 3) is represented by B, the third point(–4, 3) is represented by C and the fourth point (3, –5) is represented byD.

Distance between two points: Let P(x1, y1) Q (x2, y2) are the points in the co-ordinate plane. Let us draw a line  through P. Let R be the point of intersection of the perpendicular from Q to the line l.Then ΔPQRis a right-angled triangle. formula holds for points in all quadrants.

Also, the distance of a point P(x, y) from the origin O(0,0) is

Example: Find the distance between the points (3,4) and (6, –3).

Solution: The given points are A(3, 4) and B(6,–3). Here,  and

Therefore,Distance = units.

Example: Do the points (3, 2), (–2,–3) and (2,3) form a triangle? If so, name the type of triangle formed.

Solution: Let us apply the distance formula to find the distance PQ, QR and PR, where P(3, 2), Q(–2, –3) and R(2,3) are the given points.

We have PQ =

Since the sum of any two of these distance is greater than the third distance, therefore, the points P, Q and R form a triangle.

Also, , (50 + 2 = 52) so by the converse of Pythagoras theorem, we have ∠P = 900.

Therefore, ΔPQRis a right triangle.

Section formula: The Section Formula for the coordinates P(x, y) of a point P which divides internally the line segment joining the points  and  in the ratio as follows:

Note that, here, PA : PB = ,

However, if P does not lie between A and B but lies on the same line AB, outside the line segment AB, and PA : PB  = , we say that p divides externally the line segment joining points A and B.

Let there be two points P(x1, y1) and Q(x2,y2) in a plane. Now a point R(x, y) which divides the line segment PQ in the ratio of m1 : m2. Thus, the co-ordinates of the point R are

Coordinates of the mid-point of a line segment: If R is the mid point of PQ, then,
(as R divides PQ in the ratio 1 : 1)

Coordinates of the midpoint are

Example: Find the co-ordinates of a point which divides the line segment joining each of the following points in the given ratio:

(a) (2,3) and (7, 8) in the ratio 2 : 3 internally.

(b) (–1,4) and (0, –3) in the ratio 1 : 4 internally.

Solution: (a) Let A(2, 3) and B(7, 8) be the given points. Let P(x, y) divide AB in the ratio
2 : 3 internally. Using section formula, we have


∴ P(4,5) divides AB in the ratio 2 : 3 internally.

(b)Let A(–1, 4) and B(0, –3) be the given points. Let P(x, y) divide AB in the ratio 1 : 4 internally. Using section formula, we have


 divides AB in the ratio 1 : 4 internally.

Example: Find the mid-point of the line segment joining two points (3, 4) and (5, 12).

Solution: Let A(3, 4) and B(5, 12) be the given points. Let C(x, y) be the mid-point of AB. Using mid-point formula, we have

∴ C(4,8) are the co-ordinates of the mid-point of the line segment joining the two points.

Prove by the use of coordinate geometry that the mid-point of the hypotenuse of a right angles triangle is equidistant from its vertices.


Proof: Let ΔAOB is a right angled triangle right angled at O. Let AB be the hypotenuse and C be the mid-point of AB. We select the rectangular system of axes as shown in the figure.

OA = a, OB = b

Therefore, C is equidistant from the three vertices of ΔAOB.