Internal division with section formula
If point P (lying on AB) divides the line segment AB joining the points
A
(
x
1
,
y
1
)
{\displaystyle \mathrm {A} (x_{1},y_{1})}
and
B
(
x
2
,
y
2
)
{\displaystyle \mathrm {B} (x_{2},y_{2})}
in the ratio m:n, then
P
=
(
m
x
2
+
n
x
1
m
+
n
,
m
y
2
+
n
y
1
m
+
n
)
{\displaystyle P=\left({\frac {mx_{2}+nx_{1}}{m+n}},{\frac {my_{2}+ny_{1}}{m+n}}\right)}
[ 6]
The ratio m:n can also be written as
m
/
n
:
1
{\displaystyle m/n:1}
, or
k
:
1
{\displaystyle k:1}
, where
k
=
m
/
n
{\displaystyle k=m/n}
. So, the coordinates of point
P
{\displaystyle P}
dividing the line segment joining the points
A
(
x
1
,
y
1
)
{\displaystyle \mathrm {A} (x_{1},y_{1})}
and
B
(
x
2
,
y
2
)
{\displaystyle \mathrm {B} (x_{2},y_{2})}
are:
(
m
x
2
+
n
x
1
m
+
n
,
m
y
2
+
n
y
1
m
+
n
)
{\displaystyle \left({\frac {mx_{2}+nx_{1}}{m+n}},{\frac {my_{2}+ny_{1}}{m+n}}\right)}
=
(
m
n
x
2
+
x
1
m
n
+
1
,
m
n
y
2
+
y
1
m
n
+
1
)
{\displaystyle =\left({\frac {{\frac {m}{n}}x_{2}+x_{1}}{{\frac {m}{n}}+1}},{\frac {{\frac {m}{n}}y_{2}+y_{1}}{{\frac {m}{n}}+1}}\right)}
=
(
k
x
2
+
x
1
k
+
1
,
k
y
2
+
y
1
k
+
1
)
{\displaystyle =\left({\frac {kx_{2}+x_{1}}{k+1}},{\frac {ky_{2}+y_{1}}{k+1}}\right)}
[ 4] [ 5]
Similarly, the ratio can also be written as
k
:
(
1
−
k
)
{\displaystyle k:(1-k)}
, and the coordinates of P are
(
(
1
−
k
)
x
1
+
k
x
2
,
(
1
−
k
)
y
1
+
k
y
2
)
{\displaystyle ((1-k)x_{1}+kx_{2},(1-k)y_{1}+ky_{2})}
.[ 1]
Triangles
P
A
Q
∼
B
P
C
{\displaystyle PAQ\sim BPC}
.
A
P
B
P
=
A
Q
C
P
=
P
Q
B
C
m
n
=
x
−
x
1
x
2
−
x
=
y
−
y
1
y
2
−
y
m
x
2
−
m
x
=
n
x
−
n
x
1
,
m
y
2
−
m
y
=
n
y
−
n
y
1
m
x
+
n
x
=
m
x
2
+
n
x
1
,
m
y
+
n
y
=
m
y
2
+
n
y
1
(
m
+
n
)
x
=
m
x
2
+
n
x
1
,
(
m
+
n
)
y
=
m
y
2
+
n
y
1
x
=
m
x
2
+
n
x
1
m
+
n
,
y
=
m
y
2
+
n
y
1
m
+
n
{\displaystyle {\begin{aligned}{\frac {AP}{BP}}={\frac {AQ}{CP}}={\frac {PQ}{BC}}\\{\frac {m}{n}}={\frac {x-x_{1}}{x_{2}-x}}={\frac {y-y_{1}}{y_{2}-y}}\\mx_{2}-mx=nx-nx_{1},my_{2}-my=ny-ny_{1}\\mx+nx=mx_{2}+nx_{1},my+ny=my_{2}+ny_{1}\\(m+n)x=mx_{2}+nx_{1},(m+n)y=my_{2}+ny_{1}\\x={\frac {mx_{2}+nx_{1}}{m+n}},y={\frac {my_{2}+ny_{1}}{m+n}}\\\end{aligned}}}
External division with section formula
If a point P (lying on the extension of AB) divides AB in the ratio m:n then
P
=
(
m
x
2
−
n
x
1
m
−
n
,
m
y
2
−
n
y
1
m
−
n
)
{\displaystyle P=\left({\dfrac {mx_{2}-nx_{1}}{m-n}},{\dfrac {my_{2}-ny_{1}}{m-n}}\right)}
[ 6]
Triangles
P
A
C
∼
P
B
D
{\displaystyle PAC\sim PBD}
(Let C and D be two points where A & P and B & P intersect respectively).
Therefore ∠ACP = ∠BDP
A
B
B
P
=
A
C
B
D
=
P
C
P
D
m
n
=
x
−
x
1
x
−
x
2
=
y
−
y
1
y
−
y
2
m
x
−
m
x
2
=
n
x
−
n
x
1
,
m
y
−
m
y
2
=
n
y
−
n
y
1
m
x
−
n
x
=
m
x
2
−
n
x
1
,
m
y
−
n
y
=
m
y
2
−
n
y
1
(
m
−
n
)
x
=
m
x
2
−
n
x
1
,
(
m
−
n
)
y
=
m
y
2
−
n
y
1
x
=
m
x
2
−
n
x
1
m
−
n
,
y
=
m
y
2
−
n
y
1
m
−
n
{\displaystyle {\begin{aligned}{\frac {AB}{BP}}={\frac {AC}{BD}}={\frac {PC}{PD}}\\{\frac {m}{n}}={\frac {x-x_{1}}{x-x_{2}}}={\frac {y-y_{1}}{y-y_{2}}}\\mx-mx_{2}=nx-nx_{1},my-my_{2}=ny-ny_{1}\\mx-nx=mx_{2}-nx_{1},my-ny=my_{2}-ny_{1}\\(m-n)x=mx_{2}-nx_{1},(m-n)y=my_{2}-ny_{1}\\x={\frac {mx_{2}-nx_{1}}{m-n}},y={\frac {my_{2}-ny_{1}}{m-n}}\\\end{aligned}}}
The midpoint of a line segment divides it internally in the ratio
1
:
1
{\textstyle 1:1}
. Applying the Section formula for internal division:[ 4] [ 5]
P
=
(
x
1
+
x
2
2
,
y
1
+
y
2
2
)
{\displaystyle P=\left({\dfrac {x_{1}+x_{2}}{2}},{\dfrac {y_{1}+y_{2}}{2}}\right)}
P
=
(
m
x
2
+
n
x
1
m
+
n
,
m
y
2
+
n
y
1
m
+
n
)
{\displaystyle P=\left({\dfrac {mx_{2}+nx_{1}}{m+n}},{\dfrac {my_{2}+ny_{1}}{m+n}}\right)}
=
(
1
⋅
x
1
+
1
⋅
x
2
1
+
1
,
1
⋅
y
1
+
1
⋅
y
2
1
+
1
)
{\displaystyle =\left({\frac {1\cdot x_{1}+1\cdot x_{2}}{1+1}},{\frac {1\cdot y_{1}+1\cdot y_{2}}{1+1}}\right)}
=
(
x
1
+
x
2
2
,
y
1
+
y
2
2
)
{\displaystyle =\left({\dfrac {x_{1}+x_{2}}{2}},{\dfrac {y_{1}+y_{2}}{2}}\right)}
Centroid of a triangle
The centroid of a triangle is the intersection of the medians and divides each median in the ratio
2
:
1
{\textstyle 2:1}
. Let the vertices of the triangle be
A
(
x
1
,
y
1
)
{\displaystyle A(x_{1},y_{1})}
,
B
(
x
2
,
y
2
)
{\textstyle B(x_{2},y_{2})}
and
C
(
x
3
,
y
3
)
{\textstyle C(x_{3},y_{3})}
. So, a median from point A will intersect BC at
(
x
2
+
x
3
2
,
y
2
+
y
3
2
)
{\textstyle \left({\frac {x_{2}+x_{3}}{2}},{\frac {y_{2}+y_{3}}{2}}\right)}
.
Using the section formula, the centroid becomes:
(
x
1
+
x
2
+
x
3
3
,
y
1
+
y
2
+
y
3
3
)
{\displaystyle \left({\frac {x_{1}+x_{2}+x_{3}}{3}},{\frac {y_{1}+y_{2}+y_{3}}{3}}\right)}
Let A and B be two points with Cartesian coordinates (x1 , y1 , z1 ) and (x2 , y2 , z2 ) and P be a point on the line through A and B. If
A
P
:
P
B
=
m
:
n
{\displaystyle AP:PB=m:n}
. Then the section formula gives the coordinates of P as
(
m
x
2
+
n
x
1
m
+
n
,
m
y
2
+
n
y
1
m
+
n
,
m
z
2
+
n
z
1
m
+
n
)
{\displaystyle \left({\frac {mx_{2}+nx_{1}}{m+n}},{\frac {my_{2}+ny_{1}}{m+n}},{\frac {mz_{2}+nz_{1}}{m+n}}\right)}
[ 7]
If, instead, P is a point on the line such that
A
P
:
P
B
=
k
:
1
−
k
{\displaystyle AP:PB=k:1-k}
, its coordinates are
(
(
1
−
k
)
x
1
+
k
x
2
,
(
1
−
k
)
y
1
+
k
y
2
,
(
1
−
k
)
z
1
+
k
z
2
)
{\displaystyle ((1-k)x_{1}+kx_{2},(1-k)y_{1}+ky_{2},(1-k)z_{1}+kz_{2})}
.[ 7]
^ a b c Clapham, Christopher; Nicholson, James (2014-09-18), "section formulae" , The Concise Oxford Dictionary of Mathematics , Oxford University Press, doi :10.1093/acref/9780199679591.001.0001 , ISBN 978-0-19-967959-1 , retrieved 2020-10-30
^ "Section Formula | Brilliant Math & Science Wiki" . brilliant.org . Retrieved 2020-10-16 .
^ https://ncert.nic.in/ncerts/l/jemh107.pdf [bare URL PDF ]
^ a b c Aggarwal, R.S. Secondary School Mathematics for Class 10 . Bharti Bhawan Publishers & Distributors (1 January 2020). ISBN 978-9388704519 .
^ a b c Sharma, R.D. Mathematics for Class 10 . Dhanpat Rai Publication (1 January 2020). ISBN 978-8194192640 .
^ a b Loney, S L. The Elements of Coordinate Geometry (Part-1) .
^ a b Clapham, Christopher; Nicholson, James (2014-09-18), "section formulae" , The Concise Oxford Dictionary of Mathematics , Oxford University Press, doi :10.1093/acref/9780199679591.001.0001 , ISBN 978-0-19-967959-1 , retrieved 2020-10-30
^ a b https://ncert.nic.in/ncerts/l/leep210.pdf [bare URL PDF ]