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Section formula

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Geometric formula for finding the ratio in which a line segment is divided by a point
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In coordinate geometry, the Section formula is a formula used to find the ratio in which a line segment is divided by a point internally or externally. It is used to find out the centroid, incenter and excenters of a triangle. In physics, it is used to find the center of mass of systems, equilibrium points, etc.

Internal Divisions

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)}

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)}

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})} .

Proof

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 Divisions

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)}

Proof

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}}}


Midpoint formula

Main article: Midpoint

The midpoint of a line segment divides it internally in the ratio 1 : 1 {\textstyle 1:1} . Applying the Section formula for internal division:

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)}

Derivation

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

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)}

In 3-Dimensions

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)}

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})} .

In vectors

The position vector of a point P dividing the line segment joining the points A and B whose position vectors are a → {\displaystyle {\vec {a}}} and b → {\displaystyle {\vec {b}}}

  1. in the ratio m : n {\displaystyle m:n} internally, is given by n a → + m b → m + n {\displaystyle {\frac {n{\vec {a}}+m{\vec {b}}}{m+n}}}
  2. in the ratio m : n {\displaystyle m:n} externally, is given by m b → − n a → m − n {\displaystyle {\frac {m{\vec {b}}-n{\vec {a}}}{m-n}}}

See also

References

  1. ^ 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
  2. "Section Formula | Brilliant Math & Science Wiki". brilliant.org. Retrieved 2020-10-16.
  3. https://ncert.nic.in/ncerts/l/jemh107.pdf
  4. ^ Aggarwal, R.S. Secondary School Mathematics for Class 10. Bharti Bhawan Publishers & Distributors (1 January 2020). ISBN 978-9388704519.
  5. ^ Sharma, R.D. Mathematics for Class 10. Dhanpat Rai Publication (1 January 2020). ISBN 978-8194192640.
  6. ^ Loney, S L. The Elements of Coordinate Geometry (Part-1).
  7. ^ 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
  8. ^ https://ncert.nic.in/ncerts/l/leep210.pdf

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