Section Formula

Section formula

Section formula is a formula to calculate the coordinates of any point P(x, y) which divides the line joining the points A(x1, y1) and B(x2, y2) in the ratio of m1:m2 internally or externally. Here are those formulas:


(i)  If P(x, y) divides AB internally in the ratio of m1:m2, then the section formula for internal division is given by,

 

P(x, y) = (("m" _"1"  "x" _"2"  "+ " "m" _"2"  "x" _"1" )/("m" _"1"  "+" 〖" m" 〗_"2"  ),("m" _"1"  "y" _"2"  "+" 〖" m" 〗_"2"  "y" _"1" )/("m" _"1"  "+" 〖" m" 〗_"2"  ))


(ii) If P(x, y) divides AB externally in the ratio of m1:m2, then the section formula for external division is given by,

 

P(x, y) = (("m" _"1"  "x" _"2"   –〖"  m" 〗_"2"  "x" _"1" )/("m" _"1" – "m" _"2"  ),("m" _"1"  "y" _"2" –" " "m" _"2"  "y" _"1" )/("m" _"1" –" " "m" _"2"  ))


 

Derivation of Section Formula


Internal Division


Let A(x1, y1) and B(x2, y2) be given two points. Let the point P(x, y) divides the line joining AB internally in the ratio of m1:m2. Then, AP:PB = m1:m2

Draw perpendiculars AL, PN and BM from A, P and B respectively to the x-axis. Then, OL = x1, ON = x, OM = x2, AL = y1, PN = y and BM = y2. Again draw perpendicular AQ and PR from A and P to the line segments PN and BM respectively.


Figure: section formula for internal division


Then,

AQ = LN = ON – OL = x – x1

PR = NM = OM – ON = x2 – x

PQ = PN – QN = PN – AL = y – y1

BR = BM – RM = BM – PN = y2 – y


Now, from the figure, In ΔPQA and ΔBRP,


i.     PQA = BRP ---> both right angles

ii.    QAP = RPB ---> corresponding angles

iii.  APQ = PBR ---> corresponding angles


DPQA ~ DBRP ------> by AAA axiom


∴ "AP" /"PB"  "= "  "AQ" /"PR"  "= "  "QP" /"RB"  ----> corresponding sides of similar triangles	 i.e.	 "m" _"1" /"m" _"2"   "="  ("x - " "x" _"1" )/("x" _"2"  "- x" ) "="  ("y - " "y" _"1" )/("y" _"2"  "- y" ) Now, taking first and second term, 	"m" _"1" /"m" _"2"   "= "  ("x - " "x" _"1" )/("x" _"2"  "- x" ) or,	m2(x – x1) = m1(x2 – x) or,	m2x – m2x1 = m1x2 – m1x or,	m1x + m2x = m1x2 + m2x1 or,	x(m1 + m2) = m1x2 + m2x1 or,	x = ("m" _"1"  "x" _"2"  "+ " "m" _"2"  "x" _"1" )/("m" _"1"  "+ " "m" _"2"  ) Again, taking first and third term,  	"m" _"1" /"m" _"2"   "= "  ("y- " "y" _"1" )/("y" _"2"  "-y" ) or,	m2(y – y1) = m1(y2 – y) or,	m2y – m2y1 = m1y2 – m1y or,	m2y + m1y = m1y2 + m2y1 or,	y(m2 + m1) = m1y2 + m2y1 or,	y = ("m" _"1"  "y" _"2"  "+" "m" _"2"  "y" _"1" )/("m" _"1"  "+" "m" _"2"  ) ∴ P(x, y) = (("m" _"1"  "x" _"2"  "+" "m" _"2"  "x" _"1" )/("m" _"1"  "+" "m" _"2"  ) ","  ("m" _"1"  "y" _"2"  "+" "m" _"2"  "y" _"1" )/("m" _"1"  "+" "m" _"2"  ))

External Division


Let A(x1, y1) and B(x2, y2) be given two points. Let the point P(x, y) divides the line joining AB externally in the ratio of m1:m2. Then, AP:PB = m1:m2

Draw perpendiculars AL, BM and PN from A, B and P respectively to the x-axis. Then, OL = x1, OM = x2, ON = x, AL = y1, BM = y2 and PN = y. Again draw perpendiculars AQ from A to PN  and PR from P to BM produced.


Figure: section formula for external division


Then,

AQ = LN = ON – OL = x – x1

PR = NM = ON – OM = x – x2

PQ = PN – QN = PN – AL = y – y1

BR = RM – BM = PN – BM = y – y2


In ΔPQA and ΔBPR,


i.     PQA = BRP ---> both right angles

ii.    QAP = RPB ---> corresponding angles

iii.  APQ = PBR ---> corresponding angles


DPQA ~ DBRP ------> by AAA axiom


∴ PQA ~ BRP ------> by AAA axiom ∴ "AP" /"PB"  "= "  "AQ" /"PR"  "= "  "QP" /"RB"   ----> corresponding sides of similar triangles	 i.e.	 "m" _"1" /"m" _"2"   "= "  ("x - " "x" _"1" )/("x - " "x" _"2"  ) "= "  ("y - " "y" _"1" )/("y -" 〖" y" 〗_"2"  ) Now, taking first and second term, 	"m" _"1" /"m" _"2"   "= "  ("x - " "x" _"1" )/("x - " "x" _"2"  ) or,	m1(x – x2) = m2(x – x1) or,	m1x – m1x2 = m2x – m2x1 or,	m1x – m2x = m1x2 – m2x1  or,	x(m1 – m2) = m1x2 – m2x1 or,	x = ("m" _"1"  "x" _"2"  "- " "m" _"2"  "x" _"1" )/("m" _"1"  "-" 〖" m" 〗_"2"  ) Again, taking first and third term,  	"m" _"1" /"m" _"2"   "= "  ("y – " "y" _"1" )/("y – " "y" _"2"  ) or,	m1(y – y2) = m2(y – y1) or,	m1y – m1y2 = m2y – m2y1 or,	m1y – m2y = m1y2 – m2y1  or,	y(m1 – m2) = m1y2 – m2y1 or,	y = ("m" _"1"  "y" _"2"  "-" 〖" m" 〗_"2"  "y" _"1" )/("m" _"1"  "- " "m" _"2"  )  ∴ P(x, y) = (("m" _"1"  "x" _"2"  "- " "m" _"2"  "x" _"1" )/("m" _"1"  "-" 〖" m" 〗_"2"  ) ","  ("m" _"1"  "y" _"2"  "- " "m" _"2"  "y" _"1" )/("m" _"1"  "-" 〖" m" 〗_"2"  ))


Note: The coordinates in external division are obtained from those of the internal division by writing – m2 for m2.


 

Mid-point Formula


If P(x, y) is the mid-point of the line joining A(x1, y1) and B(x2, y2), then the point P(x, y) divides the line AB internally in the ratio of 1:1. Therefore, by applying the section formula for internal division,

P(x, y) = (("m" _"1"  "x" _"2"  "+ " "m" _"2"  "x" _"1" )/("m" _"1"  "+" 〖" m" 〗_"2"  ) ","  ("m" _"1"  "y" _"2"  "+" 〖" m" 〗_"2"  "y" _"1" )/("m" _"1"  "+ " "m" _"2"  ))             = (("1." "x" _"2"  "+1." "x" _"1" )/"1+1"  ","  ("1." "y" _"2"  "+1." "y" _"1" )/"1+1" )             = (("x" _"1"  "+" 〖" x" 〗_"2" )/"2"  ","  ("y" _"1"  "+" 〖" y" 〗_"2" )/"2" ) ∴ Mid-point formula = (("x" _"1"  "+ " "x" _"2" )/"2"  ","  ("y" _"1"  "+" 〖" y" 〗_"2" )/"2" )

 

K-formula


If P(x, y) divides the line joining the points A(x1, y1) and B(x2, y2) in the ratio of k:1, then,


P(x, y) = (("m" _"1"  "x" _"2"  "+ " "m" _"2"  "x" _"1" )/("m" _"1"  "+" 〖" m" 〗_"2"  ) ","  ("m" _"1"  "y" _"2"  "+ " "m" _"2"  "y" _"1" )/("m" _"1"  "+" 〖" m" 〗_"2"  ))             = (("k." "x" _"2"  "+1." "x" _"1" )/"k+1"  ","  ("k." "y" _"2"  "+1." "y" _"1" )/"k+1" )             = ((〖"kx" 〗_"2"  "+" 〖" x" 〗_"1" )/"k+1"  ","  ("k" "y" _"2"  "+" 〖" y" 〗_"1" )/"k+1" )


It is a section formula in terms of k, which is called K-formula.


It is convenient to let the ratio as k:1 while solving the problems to find the ratio when the points are given. In such situations, we use the K-formula. Look at the worked out examples below.

Centroid Formula


Medians of a triangle intersect at a common point which is called centroid of a triangle. If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of a triangle ABC and G is the median, then the coordinates of centroid G is given by the following formula,


G(x, y) = (("x" _"1"  "+" 〖" x" 〗_"2"  "+" 〖" x" 〗_"3" )/"3"  ","  ("y" _"1"  "+" 〖" y" 〗_"2"  "+" 〖" y" 〗_"3" )/"3" )


Proof of Centroid Formula:


Let A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of a triangle ABC. D, E and F are the mid-points of BC, CA and AB respectively. And, medians AD, BE and CF intersect at point G.


Since D is the middle point of the side BC, then its coordinate are, D = (("x" _"2"  "+ " "x" _"3" )/"2"  ","  ("y" _"2"  "+ " "y" _"3" )/"2" ) From the plane geometry, we know that the centroid of a triangle divides the median in the ratio of 2:1. Now, if the coordinates of G are (x, y), then by section formula, x = ("2." (("x" _"2"  "+ " "x" _"3" )/"2" )"+1." "x" _"1" )/"2 + 1"  = ("x" _"1"  "+ " "x" _"2"  "+" 〖" x" 〗_"3" )/"3"  y = ("2." (("y" _"2"  "+" 〖" y" 〗_"3" )/"2" )" +1." "y" _"1" )/"2 + 1"  = ("y" _"1"  "+ " "y" _"2"  "+ " "y" _"3" )/"3"  Hence, the coordinates of centroid G is,  G(x, y) = (("x" _"1"  "+" 〖" x" 〗_"2"  "+" 〖" x" 〗_"3" )/"3"  ","  ("y" _"1"  "+" 〖" y" 〗_"2"  "+" 〖" y" 〗_"3" )/"3" ) Proved.


 

Worked Out Examples


Example 1: Find the co-ordinates of the point which divides the line joining the points (-3, -2) and (2, 8) in the ratio of 3:2 internally.


Solution:  Let, P(x, y) divides the line joining A(-3, -2) and B(2, 8) in the ratio of 3:2 internally. Here, x1 = -3	      x2 = 2	m1 = 3 y1 = -2	      y2 = 8	m2 = 2 By using formula, P(x, y) = (("m" _"1"  "x" _"2"  "+" 〖" m" 〗_"2"  "x" _"1" )/("m" _"1"  "+ " "m" _"2"  ) ","  ("m" _"1"  "y" _"2"  "+ " "m" _"2"  "y" _"1" )/("m" _"1"  "+" 〖" m" 〗_"2"  )) 	= ("3×2 + 2×-3" /"3+2"  ","  "3×8 + 2×-2" /"3+2" )            = ("6 - 6" /"5"  ","  "24 - 4" /"5" ) = ("0" /"5"  ","  "20" /"5" ) = (0,4) Hence the required point is (0,4).


Example 2: What ratio is the line joining (-3, 4) and (2, -6) divided by the point (-1, 0)?


Solution:  Let the required ratio be k:1. Then, by using  the K-formula, x = (〖"kx" 〗_"2"  "+" 〖" x" 〗_"1" )/"k+1"   i.e.	-1 = "k .2 + (-3)" /"k+1"  or,	-1 = "2k - 3" /"k+1"  or,	2k – 3 = -k – 1 or,	2k + k = -1 + 3 or,	3k = 2 or,	k = "2" /"3"  Hence the required ratio is 2:3.


Example 3: If the points A (6, 4), B (a, 2) and C (1, -1) are three points on the same straight line, find the ratio in which the point  B divides the line segment AC and also find the value of a.


Solution:  Suppose, the point B(a, 2) divides the line segment joining A(6, 4) and C(1, -1) in the ratio of m1:m2. By section formula, y co-ordinate of B is, y = ("m" _"1"  "y" _"2"  "+" 〖" m" 〗_"2"  "y" _"1" )/("m" _"1"  "+" "m" _"2"  ) or,	2 = ("m" _"1"  "×-1+ " "m" _"2"  "×4" )/("m" _"1"  "+ " "m" _"2"  ) or,	2 = (〖"-m" 〗_"1"  "+ 4" "m" _"2" )/("m" _"1"  "+" 〖" m" 〗_"2"  ) or,	2(m1 + m2) = -m1 + 4m2 or,	2m1 + 2m2 = -m1 + 4m2 or,	2m1 + m1 = 4m2 – 2m2 or,	3m1 = 2m2 or,	"m" _"1" /"m" _"2"   = "2" /"3"   Hence, the required ratio is 2:3. Now x co-ordinate of B is, 	x = ("m" _"1"  "x" _"2"  "+ " "m" _"2"  "x" _"1" )/("m" _"1"  "+ " "m" _"2"  ) i.e.	a = "2×1 + 3×6" /"2 + 3"  or,	a = "2 + 18" /"5"  or,	a = "20" /"5"  or,	a = 4 ∴ a = 4.


Example 4: Find the co-ordinates of the points of trisection of the line joining the points M (7, -3) and N (1, 3).


Solution: 


Points of trisection of a line segment joining two points are  those points which divides the given line segment into three equal parts.


Let, P(a, b) and Q(c, d) are the points of trisection of the line segment joining the points M(7, -3) and N(1, 3). So P(a, b) divides MN in the ratio of 1:2 and hence,


P(a, b) = (("m" _"1"  "x" _"2"  "+" 〖" m" 〗_"2"  "x" _"1" )/("m" _"1"  "+ " "m" _"2"  ) ","  ("m" _"1"  "y" _"2"  "+ " "m" _"2"  "y" _"1" )/("m" _"1"  "+" 〖" m" 〗_"2"  ))            = ("1×1 + 2×7" /"1 + 2"  ","  "1×3 + 2×-3" /"1 + 2" )            = ("1 + 14" /"3"  ","  "3 - 6" /"3" )            = (5, -1) And, Q(c, d) is the mid-point of PN and hence, Q(c, d) = (("x" _"1"  "+ " "x" _"2" )/"2"  ","  ("y" _"1"  "+" 〖" y" 〗_"2" )/"2" ) 	     = ("5 + 1" /"2"  ","  "-1 + 3" /"2" ) 	     = (3, 1)


Hence, the required points of trisection are P(5, -1) and Q(3, 1).

Example 5: Find the centroid of the triangle with vertices A(-2, 4), B(11, 15) and C(3, -1).


Solution:


Let G(x, y) be the centroid of the triangle ABC with vertices A(-2, 4), B(11, 15) and C(3, -1). By using formula of centroid,


x = ("x" _"1"  "+ " "x" _"2"  "+" 〖" x" 〗_"3" )/"3"  = "-2 + 11 + 3" /"3"  = 4 y = ("y" _"1"  "+ " "y" _"2"  "+ " "y" _"3" )/"3"  = "4 + 15 - 1" /"3"  = 6


Hence, the centroid of the triangle is G(4, 6).


 

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