# 13 Facts On Points In Coordinate Geometry In 2D

This is a sequential post related to Co-ordinate Geometry, specially on Points. We already discussed few Topic earlier in the post “A Complete Guide to Coordinate Geometry”. In this post we will discuss the remaining topics.

# Basic Formulae on Points in Co-ordinate Geometry in 2D:

All the basic formulae on points in Analytical Geometry are described here and for easy and quick learning at a glance about the formulae a ‘Formulae Table on Points’ with graphical explanation is presented below.

## Points section formulae |Coordinate Geometry :

#### If a point divides a line segment joining two given points at some ratio, we can use section formulae to find the coordinates of that point while the ratio at which the line segment is divided by, is given and vice versa. There is a possibility that the line segment can be divided either internally or externally by the point. When the point lies on the line segment between the two given points, Internal section formulae is used i.e.

x = mx2+nx1/(m+n)

and

y = my2+ny1/(m+n)

And when the point lies on the external part of the line segment joining the two given points, external section formulae is used i.e.

y = my2-ny1/(m-n)

Where (x , y) is supposed to be the required coordinates of the point. These are very necessary formulae to find the centroid, incenters ,circumcenter of a triangle as well as the center of mass of systems, equilibrium points etc. in physics. Must watch the short view of different types of section formulae with graphs given below in the ‘Formulae Table on Points topic no 3; case-I and case-II’.

## Mid Point Formula| coordinate Geometry:

#### If a point divides a line segment joining two given points at some ratio, we can use section formulae to find the coordinates of that point while the ratio at which the line segment is divided by, is given and vice versa. There is a possibility that the line segment can be divided either internally or externally by the point. When the point lies on the line segment between the two given points, Internal section formulae is used i.e.

x = mx2+nx1/(m+n)

and

y = my2+ny1/(m+n)

And when the point lies on the external part of the line segment joining the two given points, external section formulae is used i.e.

x = mx2-nx1/(m-n)

and

y = my2-ny1/(m-n)

Where (x , y) is supposed to be the required coordinates of the point. These are very necessary formulae to find the centroid, incenters ,circumcenter of a triangle as well as the center of mass of systems, equilibrium points etc. in physics. Must watch the short view of different types of section formulae with graphs given below in the ‘Formulae Table on Points topic no 3; case-I and case-II’.

## Mid Point Formula| coordinate Geometry:

#### It is a easy formulae derived from Internal points section formulae described above. While we need to find the midpoint of a line segment i.e coordinate of the point which is equidistant from the two given points on the line segment i.e the ratio gets 1:1 form, then this formula is required. The formula is in the form of

x=x1+x2/2

and

x=y1+y2/2

Go through the “Formulae Table on Points topic no 3- case-III’ below to get the graphical idea on this.

## Area of a triangle in Coordinate Geometry:

A triangle has three side and three vertices on the plane or in 2 dimensional field. Area of the triangle is the internal space surrounded by these three sides. The basic formula of area calculation of a triangle is (1/2 X Base X Height). In Analytical Geometry ,if the coordinates of all the three vertices are given, the area of the triangle can easily be calculated by the formula, Area of the Triangle   =|½[x1 (y2  y3 )+x2 (y3  y2)+x3 (y2-y  1)]| ,actually this can be derived from the basic formula of area of a triangle using two points distance formula in coordinate geometry. Both the cases are graphically described in the ‘Formulae Table on Points topic 4’ below.

## Collinearity of points ( Three points) |Coordinate Geometry:

Collinear means ‘being on the same line’. In geometry, if three points lie on one single line in the plane, they never can form a triangle with area other than zero i.e. if the formula of area of triangle is substituted by the coordinates of the three collinear points, the result for area of the imaginary triangle formed by those points will end up with zero only. So the formula becomes like ½[x1 (y2  y3 )+x2 (y3  y2)+x3 (y2-y  1)] =0 For more clear idea with graphical representation, go through the “Formulae Table on Points topic no 5” below.

## Centroid of a triangle| Formula :

The three medians* of a triangle always intersect at a point, located in the interior of the triangle and divides the median in the ratio 2:1 from any vertex to the midpoint of the opposite side. This point is called the centroid of the triangle. The formula to find the coordinates of the centroid is

x=x1+x2+x3/3

and

x=y1+y2+y3/3

In the “Formulae Table on Points topic no 6” below, the above subject is described graphically for better understanding and for a quick view.

## Incenter of a triangle|Formula:

It is the center of the triangle’s largest incircle which fits inside the triangle. It is also the point of intersection of the three bisectors of the interior angles of the triangle. The formula, used to find the incenter of a triangle is

x=ax1+bx2+cx3/a+b+c

and

x=ay1+by2+cy3/a+b+c

In the “Formulae Table on Points topic no 6” below, the above subject is described graphically for better understanding and for a quick view.

For easy graphical explanation the below “Formulae Table on Points topic no 7” is needed to see.

## Shifting of Origin formula| Coordinate Geometry:

We have already learned in the previous post “A Complete Guide to Coordinate Geometry” that the origin lies on the point (0,0) which is the point of intersections of the axes in the plane. we can move the origin in all the quadrants of plane in respect with the origin , which will give new set of axes through it.

For a points in the above said plane , its coordinates will change along with the new origin and axes and that can be calculated by the formula, new coordinates of a point P(x1,y1) are x1 = x- a ; y1 = y-  b where the coordinates of the new origin is (a,b). To have clear understanding on this topic its preferable to see the graphical representation below in the “Formulae Table on Points topic no 8” .

## `Formulae table on Points in Coordinate Geometry in 2D`:

### ﹡Circumcentre of a triangle :

It is the point of intersection of three perpendicular bisectors of the side of a triangle. It is also the center of a triangle’s circumcircle which only touches the vertices of the triangle.

### ﹡Medians:

Median is the line segment joining the vertex of triangle to the midpoint or the point, bisecting the opposite side of the vertex. Every triangle has three medians which always intersect each other at the centroid of the same triangle.

## Solved Problems on Points in Coordinate Geometry in 2D.

For better learning about points in 2D, one basic example is solved here step by step and for practice on your own there are more problems with answers on each formula. There must be challenging problems with solution in the next articles just after getting a basic and clear idea on the topic of points in coordinate Geometry 2D.

## Problems 1:  Calculate the distance between the two given points (1,2) and (6,-3).

Solution: We already know, the formula of the distance  between two points  (x1,y1) and (x2,y2)  is d =√ [(x2-x1)2+(y2-y1)2 ] …(1)

(See the formulae table above)   Here, we can assume that (x1,y1) ≌ (1,2) and (x2,y2) ≌ (6,-3) i.e x1=1, y1=2 and x2=6, y2 =-3 , If we put all these values in the equation (1),we get the required distance.

Therefore, the distance between the two points (1,2) and (6,-3) is

=√ [(6-1)2+(-3-2)2 ]   units

= √ [(5)2+(-5)2 ]  units

=√ [25+25 ] units

=√ [50 ]        units

=√ [2×52 ]       units

= 5√2   units (Ans.)

Note: Distance is always followed by some units.

Ans. √13 units

Ans. 5 units

Ans. √41 units

Ans. 25 units

## Problem 6: Find the distance between the two points (10,100) and (-10,100,).

Ans. 20 units

Ans. √5 units

Ans. 4 units

## Problem 9: Find the distance between the two points (2+√10, 0) and (2-√10, 0).

Ans. 2√10 units

## Problem 10: Find the distance between the two points (2+3i, 0) and (2-3i, 10).  { i=√-1 }

Ans. 8 units

## Problem 11: Find the distance between the two points (2+i, -5) and (2-i, -7).  { i=√-1 }

Ans. 0 units

## Problem 12: Find the distance between the two points (7+4i,2i) and (7-4i, 2i).  { i=√-1 }

Ans. 8i units

## Problem 13: Find the distance between the two points (√3+i, 3) and (2√3+i, 5). { i=√-1 }

Ans. √7 units

## Problem 14: Find the distance between the two points (5+√2, 3+i) and (2+√2, 7+2i). { i=√-1 }

Ans. 2√(6+2i) units

## Problems 15: Find the distance of a point (3,4) from the origin.

Solution:

We have the formula of the distance of a point from the origin,  OP=√ (x2 + y2) (See the formulae table above) So here we can assume (x,y) ≌ (3,4) i.e x=3 and y=4

Therefore, putting these values of x and y in the above equation, we get the required distance

=(32 + 42)    units

=√ (9 + 16)    units

=√ (25)         units

= 5  units

Note: Distance is always followed by some units.

Note: Distance of a point from the origin is actually the distance between the point and the point of origin i.e (0,0)

Ans. √65 units

Ans. 7 units

Ans. 5 units

Ans. 10 units

## Problem 20: Find the distance of a point (0,0) from the origin.

Ans. 0 units

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