Chapter 12.4 - Section Formula

In the previous section, we completed a discussion on distance formula. In this section, we will see section formula.

• In our earlier classes, we have seen the application of section formula in two-dimensional problems.
• Let us recall the important points related to the section formula. It can be written in 2 steps:
1. In fig.12.13 (a) below, the line segment PQ lies in the XY-plane.    
    ♦ Point R lies between P and Q.
    ♦ Point R divides the segment PQ internally in the ratio m:n
    ♦ Coordinates of P and Q are (x₁,y₁) and (x₂,y₂) respectively.
• Then the coordinates of R will be: $\left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right)$.

Fig.12.13

2. In fig.12.13 (b) above, the line segment PQ lies in the XY-plane.    
    ♦ Point R does not lie between P and Q. It is on the extension of the line PQ.
    ♦ Point R divides the segment PQ externally in the ratio m:n
    ♦ Coordinates of P and Q are (x₁,y₁) and (x₂,y₂) respectively.
• Then the coordinates of R will be:$\left(\frac{m x_2 - n x_1}{m-n}, \frac{m y_2 - n y_1}{m-n} \right)$.

---

• Now we will see the section formula when P(x₁,y₁,z₁) and Q(x₂,y₂,z₂) are two points in three-dimensional space.
    ♦ R (x,y,z) divides PQ in the ratio m:n.
    ♦ We need to derive expressions for x, y and z.
• First, we will derive an expression for z. It can be derived in 8 steps:

1. The first step is to drop perpendiculars from P, Q and R, on to the XY-plane.
• In fig.12.14 below, the XY-plane is shown in blue color.
    ♦ L is the foot of the perpendicular from P.
    ♦ M is the foot of the perpendicular from Q.
    ♦ N is the foot of the perpendicular from R.

Fig.12.14

2. Draw the line LM.
    ♦ Point N will lie on LM.
3. Through R, draw a line parallel to LM.
    ♦ Let this line meet PL (actually, it is the ‘extension of PL’) at S.
    ♦ Let this line meet QM at T.
4. The points P, Q, R, S and T are on the same plane. So we can show them separately as in fig.12.15 below:

Fig.12.15

• Consider the two triangles ΔPRS and ΔQRT.
• Let us analyze the angles of the two triangles. The analysis can be written in 5 steps:
(i) PS and QT are vertical lines because, they are drawn perpendicular to the XY-plane.
(ii) ST is a horizontal line because, it is drawn parallel to the XY-plane.
(iii) So we get: ∠PSR = ∠QTR = 90^o
(iv) Angle at R is common to both triangles.
(v) So we have two angles same in both triangles. Consequently, the third angles (∠SPR and ∠TQR) will also be the same.
5. Since angles are the same in the two triangles, they are similar triangles.
• Since they are similar triangles, we can take ratio of sides:

(i) $\frac{\text{Side opposite ∠R in ΔPRS}}{\text{Side opposite ∠R in ΔQRT}}~=~\frac{\text{SP}}{\text{TQ}}$

(ii) $\frac{\text{Side opposite ∠PSR in ΔPRS}}{\text{Side opposite ∠QTR in ΔQRT}}~=~\frac{\text{PR}}{\text{QR}}~=~\frac{m}{n}$

(iii) $\frac{\text{Side opposite ∠SPR in ΔPRS}}{\text{Side opposite ∠TQR in ΔQRT}}~=~\frac{\text{SR}}{\text{TR}}$

6. For similar triangles, the three ratios will be equal. So we get:

$\frac{\text{SP}}{\text{TQ}}~=~\frac{m}{n}~=~\frac{\text{SR}}{\text{TR}}$ 

• We need only the first two ratios. We can write:

$\frac{m}{n}~=~\frac{\text{SP}}{\text{TQ}}$

7. Now we can write the lengths in terms of 'z'. It can be done in two steps:
(i) From fig.12.14, we get:
SP = SL – PL
• S is at the same level as R.
So SL = RN = z
• Also, PL = height of P = z₁
• Therefore, SP = z-z₁
(ii) Again from Fig.12.14, we get:
TQ = QM – TM
• T is at the same level as R
So TM = RN = z
• Also, QM = height of Q = z₂
• Therefore, TQ = z₂ – z
8. Substituting for SP and TQ in (6), we get:

$\begin{array}{ll} {}&{\frac{m}{n}} & {~=~}& {\frac{z - z_1}{z_2 - z}} &{} \\ {\Rightarrow}&{m(z_2 - z)} & {~=~}& {n(z - z_1)} &{} \\ {\Rightarrow}&{m z_2 ~-~m z} & {~=~}& {n z ~-~n z_1} &{} \\ {\Rightarrow}&{m z_2 ~+~n z_1} & {~=~}& {n z ~+~ m z} &{} \\ {\Rightarrow}&{(m~+~n)z} & {~=~}& {m z_2 ~+~n z_1} &{} \\ {\Rightarrow}&{z} & {~=~}& {\frac{m z_2 + n z_1}{m+n}} &{} \\ \end{array}$

---

• So we obtained the z-coordinate of R. We did this by dropping perpendiculars from P and Q, onto the XY-plane.
• If we drop perpendiculars from P and Q, onto the XZ-plane, we will get the y-coordinate of R.
• If we drop perpendiculars from P and Q, onto the YZ-plane, we will get the x-coordinate of R.
• The perpendiculars onto the XZ-plane is shown in fig.12.16 below:

Fig.12.16

• In the above fig.12.16, we see the two similar triangles ΔPRS and ΔQRT.
• Based on the above fig., the reader may write all the steps for the y-coordinate in his/her own notebooks.
   ♦ The result will be: $y~=~\frac{m y_2 + n y_1}{m+n}$
• Similarly, the reader may draw the diagrams and write the steps for x-coordinate also.
   ♦ The result will be: $x~=~\frac{m x_2 + n x_1}{m+n}$

---

So we can write a summary:

Case 1:
   ♦ P(x₁,y₁,z₁) and Q(x₂,y₂,z₂) are two points in space.
   ♦ Point R divides the line segment PQ internally in the ratio m:n
   ♦ Then the coordinates of R will be: $\left(\frac{m x_2 + n x_1}{m+n},~\frac{m y_2 + n y_1}{m+n},~\frac{m z_2 + n z_1}{m+n} \right)$

Case 2:
   ♦ P(x₁,y₁,z₁) and Q(x₂,y₂,z₂) are two points in space.
   ♦ Point R divides the line segment PQ externally in the ratio m:n
   ♦ Then the coordinates of R will be: $\left(\frac{m x_2 - n x_1}{m-n},~\frac{m y_2 - n y_1}{m-n},~\frac{m z_2 - n z_1}{m-n} \right)$

Case 3:
   ♦ Consider case 1.
   ♦ If R is the midpoint of PQ, then m:n will be 1:1
   ♦ So the coordinates of R will be: $\left(\frac{1 × x_2 + 1 × x_1}{1+1},~\frac{1 × y_2 + 1 × y_1}{1+1},~\frac{1 × z_2 + 1 × z_1}{1+1} \right)$
   ♦ That means: If R is the midpoint, then it's coordinates will be: $\left(\frac{x_2 + x_1}{2},~\frac{y_2 + y_1}{2},~\frac{z_2 + z_1}{2} \right)$

Case 4:
   ♦ Any ratio m:n can be written as k:1
   ♦ For that, we divide both m and n by n
   ♦ $\frac{m}{n}~=~\frac{m/n}{n/n}~=~\frac{m/n}{1}~=~\frac{k}{1}$
         ✰ So we get: $k~=~\frac{m}{n}$
   ♦ So the coordinates of R will be: $\left(\frac{k × x_2 + 1 × x_1}{k+1},~\frac{k × y_2 + 1 × y_1}{k+1},~\frac{k × z_2 + 1 × z_1}{k+1} \right)$
   ♦ That means, coordinates of R are: $\left(\frac{k x_2 + x_1}{1+k},~\frac{k y_2 + y_1}{1+k},~\frac{k z_2 + z_1}{1+k} \right)$
• This case 4 is helpful in some special problems where, we need to calculate only one unknown value 'k' instead of two unknown values 'm' and 'n'.

---

Now we will see a solved example:

Solved example 12.7
Find the coordinates of point R which divides the line segment joining P(1,-2,3) and Q(3,4,-5) in the ratio 2:3 (i) internally, (ii) externally.
Solution:
Part (i): Dividing internally
1. The ratio m:n is 2:3
2. Coordinates of R will be: $\left(\frac{m x_2 + n x_1}{m+n},~\frac{m y_2 + n y_1}{m+n},~\frac{m z_2 + n z_1}{m+n} \right)$
3. Substituting the values, we get:
$\left(\frac{2 × 3 + 3 × 1}{2+3},~\frac{2 × 4 + 3 × -2}{2+3},~\frac{2 × -5 + 3 × 3}{2+3} \right)$
⇒ $\left(\frac{6 + 3}{5},~\frac{8 -6}{5},~\frac{-10 + 9}{5} \right)$
⇒ $\left(\frac{9}{5},~\frac{2}{5},~\frac{-1}{5} \right)$

Part (ii): Dividing externally
1. The ratio m:n is 2:3
2. Coordinates of R' will be: $\left(\frac{m x_2 - n x_1}{m-n},~\frac{m y_2 - n y_1}{m-n},~\frac{m z_2 - n z_1}{m-n} \right)$
3. Substituting the values, we get:
$\left(\frac{2 × 3 - 3 × 1}{2-3},~\frac{2 × 4 - 3 × -2}{2-3},~\frac{2 × -5 - 3 × 3}{2-3} \right)$
⇒ $\left(\frac{6 - 3}{-1},~\frac{8 +6}{-1},~\frac{-10 - 9}{-1} \right)$
⇒ $\left(\frac{3}{-1},~\frac{14}{-1},~\frac{-19}{-1} \right)$
⇒ (-3, -14, 19)

Check:
1. Fig.12.17 below shows the actual plot:

Fig.12.17

2. We have the coordinates of all four points:
P(1,-2,3), Q(3,4,-5), R(9/5, 2/5, -1/5) and R'(-3, -14, 19)
3. First we will check the internal division:
• Using the distance formula, we can find the lengths:
    ♦ PR = 37/9 units
    ♦ QR = 49/8 units
• Thus the ratio PR/QR = $\frac{37/9}{49/8}~=~\frac{2}{3}$
4. Next we will check the external division:
• Using the distance formula, we can find the lengths:
    ♦ PR' = 102/5 units
    ♦ QR' = 153/5 units
• Thus the ratio PR'/QR' = $\frac{102/5}{153/5}~=~\frac{2}{3}$ 

---

In the next section, we will see a few more solved examples. 

Previous

Contents

Next

Copyright©2023 Higher secondary mathematics.blogspot.com