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# Cartesian coordinate System

The French Mathematician Rene Descartes invented the cartesian system. In a Cartesian coordinates form we have a point, which shows the particular place in a segment for representation. Cartesian coordinates system consist of many coordinates which are polar, rectangular etc.

In a Cartesian coordinates we have collinear point coordinates, mid point coordinates and the equidistant point coordinates:

1) Collinear points: Collinear points is a point where the three or more points lies on same line.
2) Midpoint: Mid point is a halfway point where the line segment divided into two equal parts.
3) Equidistant point: In a line segment a point is equal length from other points which are in congruent then the point are equidistant point.

 Related Calculators Cartesian Coordinates Calculator Cartesian to Polar Calculator Convert Polar to Cartesian Calculator Linear System of Equations Solver

## Definition

A system of points on a plane or in space by their coordinates is called a Cartesian system. The co ordinate axis are the two perpendicular lines which intersect at the origin.

These two lines of the Cartesian system are called the X and Y axis:

The value of x is positive in the half-plane to the right of origin, and is negative to the left of the origin.

The value of y is positive in the upper half and negative in the lower half of the XYplane. The plane is therefore divided into four quadrants

The four quadrants are the roman numerals I, II, III, and IV.

The values of X and Y in the four quadrants are listed below:

I

> 0

> 0

II

< 0

> 0

III

< 0

< 0

IV

> 0

< 0

## Cartesian System in three Dimension

The three dimensional cartesian system provides the physical dimensions of height, width, and length of something.

This system is represented using the right-hand rule, and is also called the right-handed coordinate system.

When the thumb, index and the middle finger of the right hand are at perpendicular to each other, we have X, Y, and Z axes respectively, and the points of the cartesian system are seperated by commas, as in (2, 4, 7)

To make a figure larger or smaller we need to multiply the Cartesian coordinates of every point by the same positive number. The reflection of a point is got by changing their signs in the cartesian system.

## Properties of Cartesian Systems

In coordinate geometry, the tools of algebra are used in studying geometry by establishing 1-1 correspondence between the points in a plane and the ordered pairs of real numbers:

1) If P($x_1,y_1$) and Q ($x_2, y_2$) be any two points in the plane, then PQ = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

2) If A($x_1,y_1$), B($x_2, y_2$) and C($x_3, y_3$) be the vertices of a triangle, then area, D, of triangle ABC is given by:

3) $\triangle$ = $\frac{1}{2}$ $[(x_1 y_2 - x_2y_1) +(x_2y_3 - x_3 y_2) + (x_3y_1 - x_1 y_3)]$

4) The points A, B and C are collinear if and only if area of triangle ABC is zero.

5) A point R is said to divide PQ internally in the ratio m:n if R is within PQ and $\frac{PR}{RQ}$ = $\frac{m}{n}$

6) If R(x, y) divides the join of P($x_1,y_1$) and Q($x_2, y_2$) internally in the ratio m:n, then x = $\frac{m x_2 + n x_1} {m+n}$, y= $\frac{m y_2 + n y_1}{m+n}$

7) A point R is said to divide PQ externally in the ratio m:n, where m ยน n, if R is outside  PQ and $\frac{PR}{RQ}$ = $\frac{m}{n}$

8) If R(x, y) divides the join of P($x_1,y_1$) and Q($x_2, y_2$) externally in the ratio m:n, where $m^1 n$, then  $\frac{m x_2 - n x_1} {m-n}$, y= $\frac{m y_2 - n y_1}{m- n}$

9) If M(x, y) is the midpoint of the join of P($x_1,y_1$) and Q($x_2, y_2$), then X= $\frac{x_1 + x_2}{2}$ , y = $\frac{y_1 + y_2}{2}$

10) If A($x_1,y_1$), B($x_2, y_2$) and C($x_3, y_3$) are the vertices of a triangle, then

Centroid = $\frac{x_1+x_2+x_3}{3}$, $\frac{y_1 + y_2 + y_3}{3}$

11) Incentre =  $\frac{ax_1 +bx_2 + cx_3}{a+b+c}$, $\frac{ay_1+ by_2 + cy_3}{a+b+c}$ where a = BC, b = CA, c = AB

12) When a point moves so as to always satisfy a given condition or conditions, the path it traces out is called its locus under these conditions.

13) If P(x, y) be any sample point on the locus, then an equation involving x and y which is satisfied by each point on the locus and such that each point satisfying the equation is on the locus is called the equation of the locus.

14) The slope of a non-vertical line is defined as the tangent of the angle which the line makes with the positive direction of the x-axis.

15) If a non-vertical line passes through two distinct points ($x_1,y_1$) and ($x_2,y_2$), then the slope, m, of the line is given by m= $\frac{y_2 - y_1}{x_2 - x_1}$

16) The slope of a horizontal line is zero.

17) The slope of a line making equal intercepts on the axes is '-1'.

18) The slope of a line equally inclined to the axes is either '1' or '-1'.

19) The slope of the line ax + by + c = 0, $b \neq$ 0 is equal to - $\frac{a}{b}$

20) Two non-vertical lines are parallel if and only if their slopes are equal.

21) Two non-vertical lines are perpendicular if and only if the product of their slopes is '-1'.

## Examples

Given below there are few example problems on Cartesian system

Example 1: Find the Cartesian coordinates distance between the points A(1,6) and B (4,3).

Solution: Let "d" will be the distance between A and B. ($x_1,y_1$)= (1,6), ($x_2,y_2$)= (4,3)

Then d (A, B) = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

= $\sqrt{(4-1)^2 +(3-6)^2}$

= $\sqrt{3^2+(-3)^2}$

= $\sqrt{9+9}$

=$\sqrt {18}$

=3$\sqrt{2}$
Example 2: Determine the mid point coordinates of a line segment joining given points A(5,8) and B(-1,-5)

Solution: The required mid point is

Formula =  $\frac{(x_1+x_2)}{(2)}$,$\frac{(y_1+ y_2)}{(2)}$ here, $(x_1, y_1)$ = (5,8),($x_2, y_2$) = (-1,-4)

= $\frac{(5-1)}{(2)}$ $\frac{(8-(-5))}{(2)}$

=($\frac{4}{2}$) ($\frac{13}{2}$)

= (2, 6.5)
Example 3: Find the centroid coordinates of a triangle whose vertices's points are given (0, -3), (4,6) and (2,3).

Solution: ($x_1 y_1$) = (0,-3), ($x_2, y_2$) = (4,6), ($x_3, y_3$) =(2,3)

Formula for centroid= $\frac{(x_1+ x_2+ x_3)}{(3)}$ ,$\frac{( y_1+ y_2+ y_3)}{(3)}$

The centroid of the triangle = $\frac{(0+4+2)}{3}$,$\frac{(-3+6+3)}{3}$

=$\frac{6}{3}$,$\frac{6}{3}$

=(2,2)
Example 4: In a parallelogram ABCD, the Co-ordinates of the vertices are A(-8, -4), B(6, -4), C(x, y), D(-3, 3). Find the Coordinates of the point C by applying mid point formula. [3 Mark]

Solution:

The diagonals in a parallelogram bisect each other. Let the point of bisection be O.

Then by mid point formula,

O is the midpoint of BD = ($\frac{(6-3)}{2}$, $\frac{(3-4)}{2}$)

= ($\frac{3}{2}$, $\frac{-1}{2}$)

O is also the midpoint of diagonal CA =($\frac{(x-8)}{2}$, $\frac{(y-4)}{2}$)

But Coordinates of O are ($\frac{3}{2}$, $\frac{1}{2}$)

= ($\frac{3}{2}$, $\frac{-1}{2}$) = ($\frac{(x-8)}{2}$,$\frac{(y-4)}{2}$)

$\rightarrow$ $\frac{(x-8)}{2}$ = $\frac{3}{2}$ , $\frac{(y-4)}{2}$ = $\frac{-1}{2}$

$\rightarrow$ x-8 = 3, y-4 =-1

$\rightarrow$ x=11 , y= 3

Coordinates of C(11, 3).

Example 5: Points A and B have vertices (7, -2) and (a, b) respectively. The Coordinates of the mid point are (4, -3). Find the values of a and b. [2 Mark]

Solution:

By mid point formula, Coordinates of mid point can be calculated by,

(x,y) = ($\frac{x_1 + x_2}{2}$, $\frac{y_1 + y_2}{2}$)

$\rightarrow$ (4, -3) = ($\frac{(7+a)}{2}$, $\frac{(-2+b)}{2}$ )

$\rightarrow$ 4= $\frac{(7+a)}{2}$, -3 = $\frac{(-2+b)}{2}$

$\rightarrow$ 8= 7+a, -6 = -2+b

$\rightarrow$ 1=a, -4=b

Hence a = 1, b = -4

Coordinates of B(1, -4).

## Practice Problems

Problem 1: Find the slope of the line $\frac{x}{2} + \frac{y}{2} = 1$.