Area of Triangle and Collinearity

I dedicate this post to area of triangle drawn over Cartesian plane means triangle having vertices as coordinates, to find their area.

Area of Triangle in coordinate geometry

I guess you have studied " To find the area Of Triangle" with formula 

Ar (∆ ABC) = ½ × Base × Height

no doubt, this is one of the best and easy method to find area of any triangle If we have proper values of dimensions.

But if the triangle drawn over Cartesian plane, then triangle having all the three vertices as
coordinates A (X,Y), B (x, y) & C (x', y')

then area of this triangle can be formulated as 

Ar (∆ABC) =  ½ [ X (y - y') + x ( y' - Y) + x' ( Y - y) ]

for an example:-

if,  A (2,4), B (5,-6), C( -3,-5)

Then
Ar (∆ ABC) = ½ [ 2{-6 - (-5)} + 5{-5 -4} + (-3){4 -(-6)} ]

                 = ½[ 2(-6+5) + 5(-9) -3(4+6)]
                 = ½ [ 2( -1) -45 -30 ]
                 = ½ [ -2 -75 ]
                 = ½ [- 77]
                 = [-77/2]
     absolute value will be taken as area can not be in negative i.e.

 Ar (∆ABC) = 77/2   sq.unit

same area can be evaluated from the basic formula i.e.
area of Triangle =½ × base × height
base and height can be calculated from distance formula.

for Distance formula check out Distance formula- How to apply

Collinearity ( Points on same line):-

I have explained collinearity earlier in my other post ( you can read it here Collinearity)

but there is one more way to prove, for any three coordinates, are collinear or not.

For this we can use the same formula which we have used for finding area of any triangle, but the difference is this time the formula will placed equal to zero i.e.

Three vertices as  coordinates A (X,Y), B (x, y) & C (x', y')

then for collinearity the value of above formula i.e.

         ½ [ X (y - y') + x ( y' - Y) + x' ( Y - y) ] should be equal to zero i.e.

 ½ [ X (y - y') + x ( y' - Y) + x' ( Y - y) ]  = 0

maybe some curious minds can rise the question " why this happens - formula become equal to zero"

for those I have my own perspective which I think to share i.e.

when we are looking to find area of Triangle, one simply puts the respective values into formula and can surely gets the right answer,

but when we talk about Collinearity ( points on same line) which means those particular points aren't occupying​ area that's why we'll have to put above mentioned formula equal to zero or simply if we'll have to prove for some points whether they are collinear or not, we'll have to put the values into formula and if calculation comes to zero that means points are collinear.

Okay..!!

so this is all for today,

I hope you've got a good knowledge. Now over to you, Share if you want to add something that I'm missing let us know by commenting I would like to hear from you.
keep learning.

Section Formula

      Today I'm going to explain section formula using coordinates on Cartesian plane.

But before this, please go through my previous tutorial for basics of Cartesian plane and coordinate geometry i.e.

•  Introduction to co-ordinate geometry

and

• Distance formula and Mid Point Formula

Section Formula

If a point {say (X, Y)} is dividing any line segment [ having points as (x, y) and (x', y')]

in ratio M : N then point (X, Y) can be evaluate by section formula i.e.

X =  [ M x' + N x ]  / [ M+ N ]

Y =  [ M y' + N y ] /  [ M + N ]

with this formula we can have exact location of points lying in between two points and this formula is also used for trisection and four section and onwards.

For an example:-

• Bisection :-

Q- if A (2, 4) and B (-7, 3) and a point P (x, y) is dividing the line into ratio 2 : 3 then find the value of P ???

Sol. for P (x, y) 

x = [ 2×(-7) + 3×2 ] / (2+3)

x = [ -14 + 6] / 5

x = (-8) / 5

x= -8/5

similarly

y= [ 2× 3 + 3 × 4] / (2 + 3)

y= [6 + 12 ] / 5

y= 18 / 5

so point P is 

P ( -8/5, 18/5).

• Trisection:-

If two points P(x, y) and Q (x', y') are trisecting ( points dividing line into three equal parts ) line segment AB with points 

A (6, 5) and B (4, 6)  then these points will be evaluated by method given below :-

{keep in mind on line segment P is lying nearby point A and at left of Q}

for point P ratio will be 1 : 2 

then 

x = [1×4 + 2×6] / (1+2)

x = 16/3

and 

y = [1×6 + 2×5] / (1+2)

y = 16/3

hence point P is 

P (16/3, 16/3)

similarly we can evaluate point Q but this we will take ratio as 2 : 1.

• To find Ratio:-

now this is slightly converse of above problems,

what if ratio is not given???

yes...!

in spite of the ratio is not given there should always be given the respective point for which the ratio is being considered.

so I'm going to clear this also 

   Q. Find by what ratio point P (-3, 5) is dividing line segment with points (-7, 6) and (3, 4) ?

  
 Sol. let the ratio be k : 1

then again we'll apply the same section formula here 

for P( -3, 5)

-3 = [ k× 3 + 1× -7]  / ( k+ 1)

by cross multiplying

-3× (k + 1) = 3k -7

-3k -3 = 3k - 7

-6k = -4

k = -4/-6

k = 2/3

i.e.

 k/1 = 2/3

k : 1 =  2: 3 

here is one advantage we can have i.e. we don't bother to solve for x and y both value to find ratio it can be evaluated from either x or y.

Now I think this explanation of Section Formula fulfil your expectations regarding section formula,
If you are finding any doubts instead please comment under the section.

Coordinate System

Hey Math's Buddies..!!

Here in this tutorial I'm going to take my previous tutorial at further stage regarding coordinate system, points and their locations, Distance between two points etc.
so let's start...

Coordinate System/Geometry

• Distance between points:-

if two points say A (x,y) & B (x', y') located somewhere in space the distance between those points can be evaluate by following formula:-

AB = √{(x-x')² + (y-y')²}

or it can be 

          √{(x'-x)² + (y'-y)²}

for example 

if points are 

A (2,-2) and  B(3,6)

then 

AB= √{(3-2)² + (6- (-2))²}

         √ { 1 + (6+2)²}

         √ { 1 + 8²}

         √ {1+ 64}

AB= √{65}.

• Collinearity :-

This means if three points say A, B, C are given on a same line then this condition can be represented with respect to distance as:-

AB + BC = AC

means if A and C are extreme points of that line and B lies in between them the  distance of AB & BC is equal to Distance of AC, which can be evaluate from distance formula mentioned above.

• Mid Point Formula :-

same as above if B point is lying exact middle of line then B point is considered​ to be mid point of line and to find the location or coordinates of that line can be evaluated from the formula given below:-

IF A (x,y) and C (x', y')  and B (X, Y)

then B can be found by 

X = (x+ x')/2

Y = (y +y')/2.

Now over to you, If you want to add some more here in this tutorial then please comment such. I would appreciate such efforts.

Introduction: Coordinate System and Cartesian Plane

          Today I begin with coordinate geometry topic which has created a boom in math history, I personally can say " I love this",

          The topic has resolved the problems of location of any thing in space through its arisen, later on with development of the topic the problems used to got better solutions and understandings.

 Coordinate System and Cartesian Plane

without wasting your time I'm going to start with the basics,

Co-ordinate geometry:- do you know if I'm writing a letter latter "A" this means this "A" has some particular location on this web page like this if anything written over blackboard or any surface, those particular writings have some locations with respect to their longitude and latitude locations.

So these latitude and longitude perspectives are known as x axis and y axis respectively.

This location is known as 2 - dimensional or 2D location.

Each point is located as (x,y) the value of x is also known as abscissa and ordinate for y values see below picture for better understanding

in spite of this there are 4 quadrants are well known and accepted which are sequenced as I,II,III & IV quadrants.

      There is an origin called (0, 0) divides quadrant into four sections among which I quadrants always stands for positive values of both axis, and II, III & IV quadrants having certain negative values of according to Cartesian plane you can have a good idea with this image 

With this explanation I think you come to know about Coordinate System/geometry.

As other math fields, this field also has a vast concepts and derivations and solutions like distance between coordinates, line, section of line segment, ratio of point on line etc which I will be taking surely in our next tutorial, until try to locate some points in your rough note book:-

(2,3)  , (4,-5), (-9,5), (-2,-8)

And Also  let me know what else you want to add here.