Coordinate Geometry-Section , Area Of Triangle

Coordinate Geometry-Section , Area Of Triangle
1.Internal division : Let A and B be the end points of a line segment . If a point P(other than A and B) lies on the line segment \(\overline {AB} \) , then we say that P divides  \(\overline {AB} \)  internally

Here, we say P divides \(\overline {AB} \) internally in the ratio AP : PB i.e., m : n
NOTE: If P divides internally thent the ratio AP : PB is positive i.e.,  and vice versa.
2. External division : Let A and B be the end points of a line segment. If a point P(other than A and B) lies on the line containing \(\overline {AB} \) (except middle of \(\overline {AB} \)), then we say that P divides  \(\overline {AB} \)externally.

Here, P divides \(\overline {AB} \) externally.
NOTE: If P divides \(\overline {AB} \) externally then the ratio AP : PB is negative and vice versa. i.e., \(\frac{{AP}}{{PB}} = \frac{m}{n} < 0\)

Coordinate Geometry-Section , Area Of Triangle

 Coordinate Geometry-Section , Area Of Triangle
Theorem 1 : If a point P(x, y) divides the line segment joining the points A(x₁,y₁) and B(x₂,y₂) in the ratio m : n then \(P\left( {x,y} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\)where
m+n\(\ne0\)

Without loss of generality, let us consider the points \(A\left( {{x_1},{y_1}} \right),P\left( {x,y} \right)\) and \(B\left( {{x_2},{y_2}} \right)\) belongs to first quadrant.
    In the figure we have 
 \(\left. \begin{gathered} AQ = x - {x_{1\,\,\,\,\,\,\,\,}} \hfill \\ PS = {x_2} - x \hfill \\ \end{gathered} \right|\) \(\begin{gathered} PQ = y - {y_1} \hfill \\ BS = {y_2} - {y_1} \hfill \\ \end{gathered} \)
Also, AP = m, BP = n
 Clearly, \(\Delta AQP \sim \Delta PSB\left( {A.A{\text{ }}similarity} \right)\)
  \(\Rightarrow \frac{{AQ}}{{PS}} = \frac{{PQ}}{{BS}} = \frac{{AP}}{{BP}}\)  
    (since if two triangles are similar their corresponding sides are proportional)
\(\therefore \frac{{x - {x_1}}}{{{x_2} - x}} = \frac{{y - {y_1}}}{{{y_2} - y}} = \frac{m}{n}\) \(\Rightarrow \frac{{x - {x_1}}}{{{x_2} - x}} = \frac{m}{n}\) and \(\frac{{y - {y_1}}}{{{y_2} - y}} = \frac{m}{n}\)
\(\left. \begin{gathered} \Rightarrow nx - n{x_1} = m{x_2} - mx \hfill \\ \Rightarrow \left( {m + n} \right)x = m{x_2} + n{x_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ \Rightarrow x = \frac{{m{x_2} + n{x_1}}}{{m + n}} \hfill \\ \end{gathered} \right|\) \(\begin{gathered} \Rightarrow ny - n{y_1} = m{y_2} - my \hfill \\ \Rightarrow \left( {m + n} \right)y = m{y_2} + n{y_1} \hfill \\ \Rightarrow y = \frac{{m{y_2} + n{y_1}}}{{m + n}} \hfill \\ \end{gathered} \)
Hence \(P\left( {x,y} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\) where \(m+n \ne 0\)
NOTE : If a point P divides \(\overline {AB} \) internally in the ratio 1 : 1 , then we say that P is the mid-point of  \(\overline {AB} \).

 Coordinate Geometry-Section , Area Of Triangle

 Coordinate Geometry-Section , Area Of Triangle
Mid Point of a Line Segment :
                        
Let A = (x₁, y₁), B = (x₂, y₂)
If P is the mid point of the line segment \(\overline {AB} \) then P divides \(\overline {AB} \)in the ratio internally. Let m : n = 1 : 1
Now,   \(P = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\frac{{m{y_2} + m{y_1}}}{{m + n}}} \right) = \left( {\frac{{1 \times {x_2} + 1 \times {x_1}}}{{1 + 1}},\frac{{1 \times {y_2} + 1 \times {y_1}}}{{1 + 1}}} \right)\) \(= \left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)\)  
i.e., the mid point of line segment joining of (x₁, y₁) and (x₂, y₂) is \(\left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)\)
Ex: Find the mid point of the line segment joining (1, 2) and (1, 4)
Sol : Let A = (x₁, y₁) = (1, 2), B =(x₂, y₂) = (1, 4)
Now, mid point of \(\overline {AB} = \left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right) = \left( {\frac{{1 + 1}}{2},\frac{{2 + 4}}{2}} \right) = \left( {1,3} \right)\)
The points of trisection of a line segment :
                
P divides \(\overline {AB} \) in the ratio 1 : 2. Q divides \(\overline {AB} \) in the ratio 2 : 1
Here P, Q are called the points of trisection of the line segment \(\overline {AB} \).
Note : The points which divide a line segment in the ratio 1 :2 and 2 : 1 are called the points of trisection of the line segment.

Coordinate Geometry-Section , Area Of Triangle

Coordinate Geometry-Section , Area Of Triangle
Theorem 2:  If a point P(x, y) divides the line segment joining A(x₁,y₁) and B(x_2,y₂)  in the ratio m : n externally, then\(P\left( {x,y} \right) = \left( {\frac{{m{x_2} - n{x_1}}}{{m - n}},\frac{{m{y_2} - n{y_1}}}{{m - n}}} \right)\) where \(m \ne n\)

Without loss of generality, let us assume the points A(x₁,y₁) and B(x_2,y₂) , and P(x, y) belongs to first quadrant.
 In the figure, we have
\(\left. \begin{gathered} AQ = x - {x_1} \hfill \\ BS = x - {x_2}\,\,\,\,\,\,\,\,\,\, \hfill \\ \end{gathered} \right|\) \(\left. \begin{gathered} PQ = y - {y_1}\,\,\,\,\,\,\, \hfill \\ PS = y - {y_2} \hfill \\ \end{gathered} \right|\) \(\begin{gathered} AP = m \hfill \\ BP = n \hfill \\ \end{gathered} \)
Clearly \(\Delta PAQ \sim \Delta PBS\left( {\because A.A{\text{ similarity}}} \right)\)
We have \(\frac{{AQ}}{{BS}} = \frac{{PQ}}{{PS}} = \frac{{AP}}{{BP}}\)
\(\left. \begin{gathered} \Rightarrow \frac{{x - {x_1}}}{{x - {x_2}}} = \frac{{y - {y_1}}}{{y - {y_2}}} = \frac{m}{n}\,\,\,\,\,\,\,\,\, \hfill \\ \Rightarrow \frac{{x - {x_1}}}{{x - {x_2}}} = \frac{m}{n} \hfill \\ \Rightarrow nx - n{x_1} = mx - m{x_2} \hfill \\ \Rightarrow \left( {m - n} \right)x = m{x_2} - n{x_1} \hfill \\ \Rightarrow x = \frac{{m{x_2} - n{x_1}}}{{m - n}} \hfill \\ \end{gathered} \right|\,\,\,\) \(\begin{gathered} \Rightarrow \frac{{y - {y_1}}}{{y - {y_2}}} = \frac{m}{n} \hfill \\ \Rightarrow ny - n{y_1} = my - m{y_2} \hfill \\ \hfill \\ \Rightarrow \left( {m - n} \right)y = m{y_2} - n{y_1} \hfill \\ \Rightarrow y = \frac{{m{y_2} - n{y_1}}}{{m - n}} \hfill \\ \end{gathered} \)
Hence \(P(x,y) = \left( {\frac{{m{x_2} - n{x_1}}}{{m - n}},\frac{{m{y_2} - n{y_1}}}{{m - n}}} \right)\) where \(m \ne n\)
NOTE: 
i))The internal division and external division formulare also known as "section formulae" or "Division formulae
ii) The Centroid (G) of a triangle divides the median in the ratio 2 : 1
iii)If \(A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right)\) and \(C\left( {{x_3},{y_3}} \right)\) are the vertices of a triangle, then its 
    Centroid \(\left( G \right) = \left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\)
iv)The point P(x, y) divides the line segment joining A(x₁, y₁) and B(x₂, y₂) in x₁-x :  x-x₂ (or) y₁-y : y-y₂
   
v) X-axis divides the line segment joining (x₁, y₁) and (x₂, y₂) in the ratio -y1:y2
vi)Y-axis divides the line segment joining(x₁, y₁) and (x₂, y₂)  in the ratio -x1:x
Ex:    Find the ratio in which (2, 1) divides the line segment joining (1, 4) & (4, -5).
Sol:    Let P(x, y) = (2, 1), given A(x₁, y₁) = (1, 4) and B(x₂, y₂) = (4, 5)
The required ratio = x-x₁ : x-x₂ = 1-2 : 2-4 = -1 : -2 = 1 : 2
The points divides the line segment in the ratio of 1:2  internally

Coordinate Geometry-Section , Area Of Triangle

Coordinate Geometry-Section , Area Of Triangle
Coordinates of different centres of a triangle :
Centroid of a triangle:
The point of concurrency of the medians of a triangle is called the centroid of the triangle. The coordinates of the centroid of the triangle with vertices.
\(\left( {{x_1},{y_1}} \right)\left( {{x_2},{y_2}} \right)and\left( {{x_3},{y_3}} \right)are\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\)
Incentre of a triangle :
The point of concurrency of the internal bisectors of the angles of a triangle is called the incentre of the triangle.

Note : 1. If \(\Delta ABC\) is equilateral incentre \(= \left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\) = centroid
2.    Incentre and centroid coincide in equilateral triangle
3.    The coordinates of the incentre of a triangle formed by the points 0(0, 0), A(a, 0) and B(0, b) are \(\left( {\frac{{ab}}{{a + b + \sqrt {{a^2} + {b^2}} }},\frac{{ab}}{{a + b + \sqrt {{a^2} + {b^2}} }}} \right)\)
Circumcentre of a triangle : The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of the sides of a triangle (i.e the lines through the mid point of a side and perpendicular to it.)

1.    Circumcentre is equidistant from the vertices of a triangle.
2.    The coordinates of the circumcentre of the triangle formed by  O(0, 0), A(x₁, y₁) & B(x₂, y₂) are \(\left( {\frac{{{y_2}\left( {x_1^2 + y_1^2} \right) - {y_1}\left( {x_2^2 + y_2^2} \right)}}{{2\left( {{x_1}{y_2} - {x_2}{y_1}} \right)}},\frac{{{x_1}\left( {x_2^2 + y_2^2} \right) - {x_2}\left( {x_1^2 + y_1^2} \right)}}{{2\left( {{x_1}{y_2} - {x_2}{y_1}} \right)}}} \right)\)

•      In acute angle triangle circumcentre lies inside the triangle.
•      In a right angle triangle circumcentre is at mid point of the hypotenuse.
•      In an obtuse angle triangle circumcentre lies out side the triangle but inside of the obtuse angle.

 Orthocentre : The altitudes of a triangle are concurrent and their point of concurrency is called Orthocentre and is generally denoted by 'H' or 'O'. 
Note : 
1.    The co-ordinates of the Orthocentre of the triangle with vertices O(0, 0), A(x₁, y₁) & B(x₂, y₂) are \(\left( {\frac{{\left( {{y_1} - {y_2}} \right)\left( {{x_1}{x_2} + {y_1}{y_2}} \right)}}{{{x_2}{y_1} - {x_1}{y_2}}},\frac{{\left( {{x_1} - {x_2}} \right)\left( {{x_1}{x_2} + {y_1}{y_2}} \right)}}{{{x_2}{y_1} - {x_1}{y_2}}}} \right)\)
2.    In acute angle triangle orthocentre lies inside the triangle in particular in an  equilateral triangle.
3.    In a right angled triangle orthocentre lies at the vertex of the right angle

Coordinate Geometry-Section , Area Of Triangle

Coordinate Geometry-Section , Area Of Triangle
Area of a triangle :
Theorem : If  \(A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right)\)and \(C\left( {{x_3},{y_3}} \right)\) are the vetices of  \(\Delta ABC,\)then its area \(\Delta = \frac{1}{2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|\)

NOTE : The area of the triangle formed with vertices 0(0,0), A(x₁,y₁,) and B(x₂,y₂) is \(\Delta = \frac{1}{2}\left| {{x_1}{y_2} - {x_2}{y_1}} \right|\) sq.units
\(\overline {AB} \) is the line segment and M is the mid-point of . If A(x₁,y₁,) and M(h,k) then B(2h-x₁,2k-y₁)
 If A(x₁,y₁,) and B(x₂,y₂) are the vertices of a and G(h,k) is the centroid, then its third vertex is  \(\left[ {3h - ({x_1} + {x_2}),3k - ({y_1} + {y_2})} \right]\)

 If ABCD is a parallelogram (or) rectangle (or) square (or) rhombus, then
            
            
 Area of ABCD = 2. Area of \(\Delta ABC\)
The median divides a triangle in to two triangles of equal area i.e.

Area of  \(\Delta ABC = 2.\,\)Area of \(\Delta ABD\)
If G is the centriod of a \(\Delta ABC\), then 

Area of \(\Delta ABC = 3.\,\) Area of \(\Delta AGB\)

If D,E,F are the mid-points of the sides of a  \(\Delta ABC\). 
    Then area of \(\Delta ABC = 4\) . Area of \(\Delta DEF\).
 If D,E,F are the mid-points of the sides of a . 
    Then the centroid of  \(\Delta ABC = \) The centroid of \(\Delta DEF\).
Points of trisection :
    The points which divides a line segment in the ratio 1 : 2 and 2 : 1 are called points of trisection.

here p divides \(\overline {AB} \) in the ratio 1 : 2 in ternally and Q divides \(\overline {AB} \) in the raito 2 : 1 internally Hence, P and Q are the points of trisections of \(\overline {AB} \).
Area of quadrilateral formed by the point A(x₁,y₁,), B(x₂, y₂), C(x₃, y₃), D(x₄, y₄) taken order is  \(\frac{1}{2}\left| \begin{gathered} {x_1}{\text{ }}{x_2}{\text{ }}{x_3}{\text{ }}{x_4}{\text{ }}{x_1} \hfill \\ {y_1}{\text{ }}{y_2}{\text{ }}{y_3}{\text{ }}{y_4}{\text{ }}{y_1} \hfill \\ \end{gathered} \right|\)