Mathematics Section Formula
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Section Formula

\color{red} ✍️ In two dimensional geometry, we have learnt how to find the coordinates of a point dividing a line segment in a given ratio internally. Now, we extend this to three dimensional geometry as follows:

\color{red} ✍️ Let the two given points be P(x_1, y_1, z_1) and Q (x_2, y_2, z_2).

\color{red} ✍️ Let the point R (x, y, z) PL, QM and RN perpendicular to the XY-plane.

Obviously PL ∥ RN ∥ QM and feet of these perpendiculars lie in a XY-plane.

The points L, M and N will lie on a line which is the intersection of the plane containing PL, RN and QM with the XY-plane.

Through the point R draw a line ST parallel to the line LM. Line ST will intersect the line LP externally at the point S and the line MQ at T, as shown in Fig 12.5.

\color{red} ✍️ Also note that quadrilaterals LNRS and NMTR are parallelograms.

\color{red} ✍️ The triangles PSR and QTR are similar. Therefore,

m/n = (PR)/(QR) = (SP)/(QT) = (SL - PL)/(QM - TM) = (NR - PL)/(QM - NR) = (z-z_1)/(z_2-z)

This implies color{fuchsia}(z = ( m z_2+ n z_1)/(m+n))

\color{red} ✍️ Similarly, by drawing perpendiculars to the XZ and YZ-planes, we get

color{fuchsia}(y = ( my_2+n y_1)/(m+n)) and color{fuchsia}(x = (m x_2+n x_1)/(m+n))

Hence, the coordinates of the point R which divides the line segment joining two points P (x_1, y_1, z_1) and Q (x_2, y_2, z_2) internally in the ratio m : n are

color{red}((( mx_2- nx_1)/(m-n) , (my_2-ny_1)/(m-n) , (mz_2-nz_1)/(m-n)))

color(blue)("Case 1 :") Coordinates of the mid-point: In case R is color(blue)("the mid-point of PQ,") then

color{fuchsia}(m : n = 1 : 1 ) so that color{fuchsia}(x = (x_1+x_2)/2 , y = (y_1+y_2)/2) and color{fuchsia}(z = (z_1+z_2)/2)

These are the coordinates of the mid point of the segment joining P( x_1 , y_1 , z_1) and Q ( x_2 , y_2 , z_2)

color(blue)("Case 2 :") The coordinates of the point R which divides PQ in the ratio k : 1 are obtained by taking k = m/n which are as given below:

color{fuchsia}((( kx_2+x_1)/(1+k) , (ky_2+y_1)/(1+k) , (kz_2+z_1)/(1+k)))

Generally, this result is used in solving problems involving a general point on the line passing through two given points.

Q 3089034817

Find the coordinates of the point which divides the line segment joining the points (1, –2, 3) and (3, 4, –5) in the ratio 2 : 3 (i) internally, and (ii) externally.

Solution:

(i) Let P (x, y, z) be the point which divides line segment joining A(1, – 2, 3) and B (3, 4, –5) internally in the ratio 2 : 3. Therefore

x = (2(3) +3(1))/(2+3) = 9/5 , y = (2(4)+3(-2))/(2+3) = 2/5 , z = (2(-5)+3(3))/(2+3) = (-1)/5

Thus, the required point is  ( 9/5 , 2/5 , (-1)/5 )

(ii) Let P (x, y, z) be the point which divides segment joining A (1, –2, 3) and B (3, 4, –5) externally in the ratio 2 : 3. Then

x = (2(3)+(-3)(1))/(2+(-3)) = -3 , y = (2(4)+(-3)(-2))/(2+(-3)) = -14 , z = (2 (-5) +(-3)(3))/(2+(-3)) = 19

Therefore, the required point is (–3, –14, 19).
Q 3039134912

Using section formula, prove that the three points (– 4, 6, 10), (2, 4, 6) and (14, 0, –2) are collinear.

Solution:

Let A (– 4, 6, 10), B (2, 4, 6) and C(14, 0, – 2) be the given points. Let the point P divides AB in the ratio k : 1. Then coordinates of the point P are

((2k-4)/(k+1) , (4k+6)/(k+1) , (6k+10)/(k+1))

Let us examine whether for some value of k, the point P coincides with point C.

On putting (2k-4)/(k+1) = 14 we get k = -3/2

when k = -3/2 , then (4k+6)/(k+1) = (4 (-3/2) +6)/(-3/2+1) = 0

and (6k+10)/(k+1) = (6(-3/2) +10)/(-3/2+1) = -2

Therefore, C (14, 0, –2) is a point which divides AB externally in the ratio 3 : 2 and is same as P.Hence A, B, C are collinear.
Q 3059245114

Find the coordinates of the centroid of the triangle whose vertices are (x_1, y_1, z_1), (x_2, y_2, z_2) and (x_3, y_3, z_3)

Solution:

Let ABC be the triangle. Let the coordinates of the vertices A, B,C be (x_1, y_1, z_1), (x_2, y_2, z_2) and (x_3, y_3, z_3), respectively. Let D be the mid-point of BC. Hence coordinates of D are

( (x_2+x_3)/2 , (y_2+y_3)/2 , (z_2+z_3)/2)

Let G be the centroid of the triangle. Therefore, it divides the median AD in the ratio 2 : 1. Hence, the coordinates of G are

( {(2 (x_1+x_2)/2)+x_1 }/(2+1) , {2((y_1+y_3)/2 )+y_1}/(2+1) , {2 ((z_2+z_3)/2)+z_1}/(2+1))

or (( x_1+x_2+x_3)/3 , (y_1+y_2+y_3)/3 , (z_1+z_2+z_3)/3)
Q 3039345212

Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, – 8) is divided by the YZ-plane.

Solution:

Let YZ-plane divides the line segment joining A (4, 8, 10) and B (6, 10, – 8) at P (x, y, z)  in the ratio k : 1. Then the coordinates of P are

((4+6k)/(k+1) , (8+10k)/(k+1) , (10-8k)/(k+1))

Since P lies on the YZ-plane, its x-coordinate is zero, i.e., (4+6k)/(k+1) = 0

or k = -2/3

Therefore, YZ-plane divides AB externally in the ratio 2 : 3.