Two Dimensional Coordinate Geometry

Coordinate Geometry is the branch of mathematics that helps us to exactly locate a given point with the help of an ordered pair of numbers. 2-dimensional geometry is the combination of geometry and algebra to solve the problems.

Two-dimensional Coordinate geometry is also said to be the study of graphs. Graphs are visual representations of our data. It can be in different forms like bar graphs, histograms, line graphs, etc.

Two-dimensional Coordinate geometry is also known as cartesian geometry.

Cartesian Coordinates (Rectangular Coordinates)

Cartesian Coordinates can be defined as a representation of a point in a plane by an ordered pair of real numbers.

Consider the below figure,

In Cartesian coordinates, the position of a point P is determined by knowing the distances from two perpendicular lines passing through the fixed point.

Some of the Important Two-Dimensional Coordinate Geometry Terms

From this figure, let us understand some important terms used in 2-dimensional geometry.

Axes of Coordinates

In the above figure OX and OY are called X-axis and Y-axis respectively. And both together are known as axes of coordinates.

Origin

The point of intersection of the axes is called the origin it is O.

Abscissa

The distance of any point on the plane from the Y-axis is called the abscissa.

Ordinate

The distance of any point on the plane from the X-axis is said to be ordinate.

Coordinate of the Origin

It has zero distance from both the axes. Hence the coordinates of the origin are (0, 0).

The axes divide the plane into four parts. These four parts are said to be quadrants.

A quadrant is ¼ th part of a plane divided by coordinate axes.

Distance Between Two Points

Let two points be A(x1,y1) and B(x2,y2).

Then Distance Formula is Given by

 $AB{\text{ }} = \sqrt {} \left[ {{{\left( {{x_2} - {\text{ }}{x_1}} \right)}^2} + {\text{ }}{{\left( {{y_2} - {\text{ }}{y_1}} \right)}^2}} \right]$

Section Formula

Let A(x1 , y1)  and B(x2, y2) be the two pints on the XY plane. And P( x, y) be any point on segment AB such that AP: PB = m:n.

Condition 1: If the point P divides segment AB internally in the ratio m:n.

Then the (x,y) coordinates of P are given by:

$\left[ {\frac{{m{x_2} + n{x_1}}}{{m + n}},\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right]$

Condition 2: If the point P divided the line segment AB externally in the ratio m:n.

Then the ( x, y) coordinates of P are given by

$\left[ {\frac{{m{x_2} - n{x_1}}}{{m - n}},\frac{{m{y_2} - n{y_1}}}{{m - n}}} \right]$

Condition 3: If the point P divided the line segment AB in the equal ratio i.e m = n or m:n = 1:1 then the ( x, y) coordinates of P are given by

 P(x,y) = $\left( {\frac{{x1 + x2}}{2},\frac{{y1 + y2}}{2}} \right)$

Area of a Triangle

The area of a triangle whose coordinates of vertices are (x1, y1) , (x2, y2) and (x3, y3)

The area of a triangle ABC whose vertices are A(x1, y1) , B(x2, y2) and C(x3, y3) is given by

 Area of ∆$= \frac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$

Note :

• To find the area of a polygon we divide it in triangles and take the numerical value of the area of each of the triangles.

• Three points A(x1, y1) , B(x2, y2) and C(x3, y3) are collinear if and only if

Area of = 0 i.e ½ [ x1( y2- y3) +  x2( y3- y1) +  x3( y1- y2) ] = 0

Solved Examples

Example 1 : P (4, 5) and Q(7, – 1) are two given points and the point R divides the line-segment PQ externally in the ratio 4:3. Find the coordinates of Y.

Solution :

Given that P(4,5)=(x1,y1),

Q(7,-1)=(x2,y2)

Point R divides the segment PQ  in the ratio 4:3,

hence m=4, n=3

Since given that the point R divides the segment externally we use the section formula for external division,

$\;R{\text{ }} = {\text{ }}\left\{ {\left[ {\frac{{\left( {m{x_2} - n{x_1}} \right)}}{{\left( {m - n} \right)}}} \right],\left[ {\frac{{\left( {m{y_2} - n{y_1}} \right)}}{{\left( {m - n} \right)}}} \right]} \right\}$

Substituting the given values,

={[((4 x 7) - (3 x 4))/(4 - 3)],[(4 x -1) - (3 x 5)/(4 - 3)]}

={(28 - 12)/1,(-4 - 15)/1}

={16,-19}

The coordinates for the point R which divides the segment PQ externally are (16,-19)

Example 2 : Find the area of the LMN whose vertices are L(3, 2), M(4, 2) and N(3, 5)

Solution:

Given that L( 3, 2) = ( x1, y1)

M( 4, 2) = ( x2, y2)

N( 3, 5) = ( x3, y3)

To calculate area of LMN we have formula

To calculate area of LMN we have formula

A =  ½  [x1 (y2- y3 )+x2 (y3 - y1)+x3(y1- y2)]

A = ½  [3(2 – 5) + 4(5 – 2) + 3(2 – 2)]

A = ½  [3 x (-3) + 4 x 3]

A = ½  [- 9 + 12]

A = ½ (3)

=  3/2  square units.

Therefore, the area of a triangle LMN  is  3/2 square units.

Quiz Time

1. Find the coordinates of point T when T divides the line segment XY where X ( 5,2) and Y( 1, 3) in the ratio 2: 4 internally.

2. Find the midpoint of segment AB where A(2, 4) and B(6, 5).

3. Find the ratio in which the point P (5, 4) divides the line segment joining points A (2,1) and B (7, 6).