Coordinate Geometry Class 10 Notes Maths Chapter 7

Terminologies related to Cartesian Plane

• Position of a point P in the Cartesian plane with respect to co-ordinate axes is represented by the ordered pair (x, y).

• 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:

• Any point on the X-axis is of the form (x, 0).
• Any point on the Y-axis is of the form (0, y).

Distance Formula

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

Note. If O is the origin, the distance of a point P(x, y) from the origin O(0, 0) is given by :

Example: Find the distance between the following points:
(i) (-1, 2) and (2, 3)
(ii) (0, 1) and (6, –1)

Solution:

(i) Let the distance between the points (-1, 2) and (2, 3) be d, then

d = √[(2 – ( –1))2 + (3 – 2)2] = √[9 + 1] = √10 units.

(ii) Let the distance between the points (0, 1) and (6, –1) be d, then

d = √[(6 – 0)2 + ( –1 – 1)2] = √[36 + 4] = √40 = 2√10 units.

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Section Formula 

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

The above formula is section formula. 

The ratio m: n can also be written as m/n  : 1 or k:1, The co-ordinates of P can also be written as P(x, y) =

Example: Find the coordinates of the point which divides the line segment joining the points (4,6) and (-5,-4) internally in the ratio 3:2.

Sol: Let P(x, y) be the point which divides the line segment joining A(4, 6) and B(-5, -4) internally in the ratio 3 : 2.

Here,

(x1, y1) = (4, 6)

(x2, y2) = (-5, -4)

m : n = 3 : 2

Using the section formula,

Coordinates of P are, 

Therefore, P(x,y) = (-7/5,0).

Finding Ratios given the Points

To find the ratio in which a given point P(x, y) divides the line segment joining A(x₁, y₁) and B(x₂, y₂),

• Assume that the ratio is k : 1
• Substitute the ratio in the section formula for any of the coordinates to get the value of k.

When x₁, x₂ and x are known, k can be calculated. The same can be calculated from the y- coordinate also.

Example: Find the ratio when point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8)?

Solution: Let the ratio be m:n.

We can write the ratio as:

m/n : 1 or k:1

Suppose (-4, 6) divide the line segment AB in k:1 ratio.

Now using the section formula, we have the following;

Thus, the required ratio is 2:7.

Mid Point of a Line Segment

The mid-point of the line segment joining the points P(x₁, y₁) and Q(x₂, y₂) is

Substituting m = 1, n = 1 in section formula we get,

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Points of Trisection

To find the points of trisection P and Q, which divides the line segment joining A(x1, y1) and B(x2, y2) into three equal parts:

i) AP : PB = 1 : 2

ii) AQ : QB = 2 : 1

Example: Find the coordinates of the points of trisection of the line segment joining the points A(2, – 2) and B(– 7, 4).

Solution: Let P and Q divide the line segment AB into three parts.

So, P and Q are the points of trisection here.

Let P divides AB in 1:2, thus by section formula, the coordinates of P are (1, 0)

Let Q divides AB in 2:1 ratio, then by section formula, the coordinates are (-4,2)

Thus, the point of trisection for line segment AB are (1,0) and (-4,2).

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Centroid of a Triangle

If A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃) are the vertices of a ΔABC, then the coordinates of its centroid(P) are given by:

Example: Find the coordinates of the centroid of a triangle whose vertices are given as (-1, -3), (2, 1) and (8, -4)

Solution: Given,

The coordinates of the vertices of a triangle are (-1, -3), (2, 1) and (8, -4)

The Centroid of a triangle is given by:

G = ((x1+x2+x3)/3, (y1+y2+y3)/3)

G = ((-1+2+8)/3, (-3+1-4)/3)

G = (9/3, -6/3)

G = (3, -2)

Therefore, the centroid of a triangle, G = (3, -2)