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Section formula is widely used to locate the coordinates of the point which divides the line segment internally or externally in a specific ratio in two-dimensional(2D) and three-dimensional (3D) planes in mathematics. In space, the coordinate system is usually used to locate the position of points. Moreover, there are three axes in a three-dimensional plane such as x-axis, y-axis and z-axis. The two main applications of the section formula are to find the centroid of the triangle and check the collinearity of three points in mathematics.
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Key Takeaways: Section Formula, Coordinate geometry, Collinearity, Coordinate system, Centroid, Segment, Plane
What is Section Formula?
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In coordinate geometry, the section formula is used to determine the coordinates of the points which divide the line segment at a specific ratio. In two dimensional, coordinate geometry has only two axes such as x-axis and y-axis. Similarly, there are three directions in a three-dimensional plane x-axis, y-axis and z-axis. Mathematically, the coordinates of any point in three dimensional plane is written as,
P(x,y,z)
Here, ‘P’ is any point in space and ‘x’,’y’,’z’ are the three directions. Moreover, when three coordinates of the point are already known, then determining the directions of any point is an easy task. In the section formula, points cut the line segment in two ways internally and externally.
Section Formula Example Diagram
Also Read:
| Related Articles | ||
|---|---|---|
| Straight line | Slope of the line | Introduction to Three-Dimensional Geometry |
| Circles | Triangles | Similarities of Triangle |
| Pythagoras Theorem | Trigonometric Identities | Construction Formula |
Derivation of Section Formula
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To determine the section formula, first we need to draw a three dimensional plane which has three points. Let us take two points like P(x1,y1,z1), Q(x2,y2,z2) which make a line. Also, there is another point R(x,y,z) which divides the line PQ into a specific ratio (m:n).
Section Formula Derivation Diagram
Draw three perpendicular PL, QM, RN to the XY- plane. Through a point R, draw a straight line SRT parallel to LNM i.e. SRT||LNM lies in the xy-plane.. Therefore, when they meet LP is produced in S and MQ in T.
The line PQR cuts all the three perpendicular lines such as PL,QM and RN. Hence, all they lie in the same plane i.e. xy-plane.
Now, angles SPR and PQR are similar.
Now, the angles PRS and TQR are similar.
∠PSR = ∠RTQ (because these are alternate angles)
∠PRS = ∠TRQ (because these are vertically opposite angles)
and
∠RPS = ∠TQR (because third angle of one triangle is equal to third angle of other triangle)
Therefore, |PS| / |TQ| = |TQ| / |RQ|
|PS| / |TQ| = m / n ….(i) (where m + n ≠ 0)
However,
PS = LS - LP
PS = NR - LP
PS = z- z1
And TQ = MQ - MT
TQ = MQ - NR
TQ = z2 - z
Put these values in above equation (i), we get,
|PS| / |TQ| = m / n
z- z1 /z2 - z = m / n
Now, cross multiply the equation, we have,
n (z- z1) = m(z2 - z)
⇒ n z -n z1 = mz2 - mz
⇒ n z + mz = mz2 +n z1
⇒ ( m + n)z = mz2 +n z1
⇒ z = mz2 + n z1 / ( m + n)
Hence, this is for the z-axis.
Similarly, for x-axis and y-axis, we have,
x= mx2 +n x1 / ( m + n)
y =my2 +n y1 / ( m + n)
We get all the values of the three axes.
Then,
The coordinates of point R are:
{mx2 +n x1 / ( m + n) my2 +n y1 / ( m + n) , mz2 +n z1 / ( m + n)}
Important Formulas
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Section Formula (Internally): When an any point {‘R(x,y,z)’} which divides the line segment joining the any two distinct points {P(x1,y1,z1),Q(x2,y2,z2)} in the specific ratio (m:n) internally then the coordinates of the point is given by,
R(x,y,z) = (mx2 +n x1 / ( m + n) my2 +n y1 / ( m + n) , mz2 +n z1 / ( m + n)).
Section Formula (Externally): When an any point {‘R(x,y,z)’} which divides the line segment joining the any two distinct points {P(x1,y1,z1),Q(x2,y2,z2)} in the specific ratio (m:n) externally (replace n with -n) then the coordinates of the given point is given by,
R(x,y,z) = {mx2 -n x1 / ( m - n) ,my2 -n y1 / ( m - n) , mz2 -n z1 / ( m - n)}
Mid point Formula: When any point R(x,y,z) cuts the line segment {P(x1,y1,z1), Q(x2,y2,z2)} in the ratio of 1:1(i.e. m=n=1), then R is the mid point. The coordinates of the mid point R are given by,
R(x,y,z) = (x2 +x1 /2 , y2 +y1 /2 , z2 +z1 /2)
Things to Remember
- Section formula is useful in locating the location of anything in the space.
- In coordinate geometry, the section formula is widely used to find the coordinates.
- When any point divides the line segment internally then all the values are positive.
- When any point divides the line segment eternally then all the values are negative.
- When any point divides the line segment at the middle of the line then the ratio is 1:1 i.e. m=n=1.
- Section Formula is widely used to find the centroid of the triangle as well as to check the collinearity of three points.
Also Read:
Sample Questions
Ques. Find the coordinates of the point, which divides the two distinct joining points (-2,3,5) and (1,-4,-6) in the ratio of 2:3 both internally and externally. [2 marks, CBSE 2011]
Ans. Given, two distinct joining points. Let us consider A=(- 2,3,5) and B= (1,- 4,- 6) . Also, assume ‘P’ is the point whose coordinates need to be determined.
Now, Section Formula Internally,
P(x,y,z) = (mx2+nx1/m+n, (my2+ny1/m+n, (mz2+nz1/m+n)
Here, x2 = 1, y2 =-4, z2 = -6, and x1 =-2, y1 =3, z1 =5.
Also, m= 2 and n=3.
Put all three values in the above section formula, we get,
P(x,y,z) = (2(1) + 3(-2)/2+3, 2(-4) + 3(3)/2+3, 2(-6) + 3(5)/2+3 )
P(x,y,z) =(2-6/5,-8+9/5 , -12+15/5)
P(x,y,z) =(-4/5,1/5 , 3/5)
Hence, the required point is P(-4/5,1/5 , 3/5).
(ii) Similarly, Section Formula Externally,
P(x,y,z) = (mx2-nx1/m-n, (my2-ny1/m-n, (mz2-nz1/m-n)
Here, x2 = 1, y2 =-4, z2 = -6, and x1 =-2, y1 =3, z1 =5.
Also, m= 2 and n=3.
Put all three values in the above section formula, we get,
P(x,y,z) = (2(1) - 3(-2)/2-3, 2(-4) - 3(3)/2-3, 2(-6) - 3(5)/2-3 )
P(x,y,z) =(2+6/-1,-8-9/-1 , -12-15/-1)
P(x,y,z) =(-8/,17 , 27)
Hence, the required point is P(-8/,17 , 27).
Ques. Determine the coordinates of the mid- point which join the two distinct points A(3,5,7) and B(-3,-3,1). [2 marks, NCERT]
Ans. Given, two distinct joining points A(3,5,7) and B(-3,-3,1). Let P(x,y,z) is the midpoint.
Now, the mid point of AB is given as,
P(x,y,z) = (x2+x1/2 , y2+y1/2 , z2+z1/2 )
P(x,y,z) = (3-3/2 , 5-3/2 , 7+1 / 2 )
P(x,y,z) = (0/2,-2/2, 8/2)
P(x,y,z) =(0,-1,4).
Hence, the required mid point is P(x,y,z) = (0,-1,4).
Ques. Determine the coordinates of the point R, which divides the two distinct joining points PQ externally in the ratio of 2:1. Also, verify that Q is the mid point of PR. [3 marks]
Ans. Given, ratio i.e. m:n is 2:1.
Let us assume that P(x1,y1,z1) and Q(x2,y2,z2) are the given points.
by using the section formula, the coordinates of the R, which divides the PQ externally is given by,
R(x,y,z) = (mx2-nx1/m-n, (my2-ny1/m-n, (mz2-nz1/m-n)
R(x,y,z) = (2x2-x1/2-1 , 2y2-y1/2-1 , 2z2-z1/2-1)
R(x,y,z) = (2x2-x1/1 , 2y2-y1/1 , 2z2-z1/1)
R(x,y,z) = (2x2-x1 , 2y2-y1 , 2z2-z1)
Hence, the coordinates of the point R is (2x2-x1 , 2y2-y1 , 2z2-z1).
Also,
Now, the mid point of PR is determined by using mid point formula,
(x2+x1/2 , y2+y1/2 , z2+z1/2 )
(x1+(2x2-x1)/2 , y1+(2y2-y1)/2 , z1+(2z2-z1)/2 )
(2x2/2 ,2y2/2, 2z2/2)
(x2 ,y2, z2) this is equal to Q.
Therefore, Q is the mid point of PR.
Ques.A point R with x coordinate 4 lies on the line segment joining the points P(2,-3,4) and Q(8,0,10). Determine the coordinates of the point R. [3 marks]
Ans. Given, P(2,-3,4), Q(8,0,10). Let us assume that point R(x,y,z) divides the line segment in the ratio of k:1.
By using the section formulaI(internally), the coordinates of the point R are given by,
[if nothing is mentioned in the statement then consider that point divides internally]
R(x,y,z) = (8k+2/k+1 ,0*k-3/k+1 ,10k+4/k+1) ….(i)
According to the question, x ordinate 4 lies on the line segment then,
8k+2/k+1 = 4
Cross multiply the terms, we get,
8k+2 =4( k+1)
⇒ 8k+2 =4 k+4
⇒ 8k-4k = 4-2
⇒ 4k = 2
⇒ k = 2/4
⇒ k =½
Now, put the value of k in the equation (i), we have,
R(x,y,z) = (8(½)+2/ ½ +1 ,0* ½ -3/ ½ +1 ,10(½)+4/ ½ +1 )
R(x,y,z) = (4+2/ 3/2 ,0-3/3/2 ,5+4/3/2)
R(x,y,z) = (6/ 3/2 ,-3/3/2 ,9/3/2)
R(x,y,z) = (6*2/ 3 ,-3*2/3 ,9*2/3)
R(x,y,z) = (12/ 3 ,-6/3 ,18/3)
R(x,y,z) = (4 ,-2 ,6)
Therefore, the coordinates of point R are (4 ,-2 ,6) .
Ques. Determine the ratio in which the plane 3x+4y-5z =1 divides the line joining the points (-2,4,6) and (3,-5,8). [3 marks, CBSE]
Ans. Let us consider P(-2,4,6) and Q (3,-5,8) are the given points. Also, the point R which divides the PQ in the ratio of k:1. Then, the coordinates of R are given by,
R(x,y,z) = ( 3k-2/k+1 ,-5k+4/k+1 ,8k-6/k+1 )
Since point R lies on the plane 3x+4y-5z =1.
Now, substitute the coordinates of R in the equation of plane, we have,
⇒3(3k-2/k+1 ) + 4(-5k+4/k+1) - 5 (8k-6/k+1) = 1
⇒ (9k-6/k+1 ) -(20k -16/k+1) -(40k -30/k+1) = 1
⇒ 9k-6 -20k +16 - 40k +30 = k+1
⇒-51k-k = -40+1
⇒ 52k =39
⇒ k = 39/52
⇒ k =¾
Therefore, Point R divides the point |PQ| in the ratio of 3:4.
Ques. Using the section formula, prove that the three points (-4,6,10),(2,4,6) and (14,0,-2) are collinear. [5 marks]
Ans. Let us assume that A (-4,6,10),B(2,4,6) and C(14,0,-2) are the given points.Also, consider that C divides the point |AB| in the ratio of k:1.
By using section formula(internally), the coordinates of C are given by,
C(14,0,-2) = ( 2k-4/k+1 ,4k+6/k+1 ,6k+10/k+1)....(i)
Now, equate the x coordinates, we get,
2k-4/k+1 = 14
⇒ 2k-4 = 14(k+1)
⇒ 2k-4 = 14k+14
⇒2k -14k = 14+4
⇒ -12k = 18
⇒k = - 18/12
⇒ k= -3/2 .
Put the value of k in equation (i) , we get,
⇒ 2(-3/2) -4/(-3/2)+1, 4(-3/2)+6/(-3/2)+1 ,6(-3/2)+10/(-3/2)+1 )
⇒(-3-4/-½ , -6+6/-½ , -9+10/-½)
⇒(-7*2/-1 , 0*2/-1, 1*2/-1)
⇒(14, 0 , -2) which is equal to point C.
Therefore, C is the point which divides the points |AB| in the ratio of 3:2.
Hence, A, B and C are collinear.
Ques. Prove that the points (3,-1,-1),(5,4,0),(2,3,-2) and (0,6,-3) are the vertices of the parallelogram. [3 marks]
Ans. Let us assume that A(3,-1,-1),B(5,4,0),C(2,3,-2) and D(0,6,-3) are the given points.
(2,3,-2) (0,6,-3)
(3,-1,-1) (5,4,0)
Parallelogram Diagram
We know that parallelogram has four vertices. Then, it contains two midpoints.
Now, the Mid point of |AC| is given by,
Take A(3,-1,-1), C(2,3,-2) vertices and apply mid point formula,
⇒(x2 +x1 /2 , y2 +y1 /2 , z2 +z1 /2 )
⇒(2+3/2,3-1/2/2, -2-1 / 2)
⇒ (5/2,2/2,-3/2)
⇒(5/2,1,-3/2)....(i)
Similarly, the mid point of |BD| is given by,
Take B(5,-4,0), D(0,6,-3) vertices and apply mid point formula,
⇒(x2 +x1 /2 , y2 +y1 /2 , z2 +z1 /2 )
⇒(0+5/2,6-4/2,-3+0/2)
⇒ (5/2, 2/2, -3/2)
⇒(5/2, 1, -3/2)....(ii)
From (i) and (ii), it is clear that mid midpoint is same for [AC] and [BD].
Thus, [AC] and [BD] have the same point, that is (5/2, 1, -3/2).
Therefore, ABCD is a parallelogram.
Ques. Find the third vertex of the triangle whose centroid is (7,-2,5) and whole other vertices are (2,6,-4) and (4,-2,3). [3 marks]
Ans. Let us assume that ABC are the three vertices of the triangle and G is the centroid. Here, A(2,6,-4),B (4,-2,3) and C(x,y,z). Also, the centroid has coordinates G(7,-2,5).
(2,6,-4)
(4,-2,3). (x,y,z) (?)
Triangle with Centroid
We know that G(7,-2,5) is the centroid of ?ABC.
∴ 2+4+x/3 = 7, 6-2+y/3 = -2 , -4+3+z/3 = 5
⇒(6+x)/3 =7 , (4+y)/3 = -2 , (-1+z)/3 = 5
⇒(6+x)= 3*7, (4+y) = -2*3 , (-1+z) = 5*3
⇒6+x = 21 , 4+y = -6 , -1+z = 15
⇒ x= 21-6 , y = -6-4 , z=15+1
⇒ x= 15 , y= -10 , z= 16
Therefore, C(15,-10,16) is the third vertex.
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