Chapter 12.5 - Solved Examples on Section Formula

In the previous section, we saw section formula. We also saw a solved example. In this section, we will see a few more solved examples.

Solved example 12.8
Using the section formula, prove that the three points A(-4,6,10), B(2,4,6) and C(14,0,-2) are collinear.
Solution:
1. Let us assume that, the third point C is collinear with A and B. Then two possibilities arise:
(i) C lies between A and B
(ii) C lies in the extension of AB
2. If the case is 1(i), then C will be dividing AB internally in some ratio k:1
3. If the case is 1(ii), then C will be dividing AB externally in some ratio k:1
4. For a point which divides AB (internally or externally) in the ratio k:1, the coordinates will be:
$\left(\frac{k x_2 + x_1}{1+k},~\frac{k y_2 + y_1}{1+k},~\frac{k z_2 + z_1}{1+k} \right)$
⇒ $\left(\frac{k × 2 + -4}{1+k},~\frac{k × 4 + 6}{1+k},~\frac{k × 6 + 10}{1+k} \right)$
⇒ $\left(\frac{2k - 4}{1+k},~\frac{4k + 6}{1+k},~\frac{6k + 10}{1+k} \right)$
5. Let us check whether the coordinates of C satisfy the conditions in (4).
The check can be done in 4 steps:
(i) Equating the x coordinates, we get:

$\begin{array}{ll} {}&{\frac{2k - 4}{1+k}} & {~=~}& {14} &{} \\ {\Rightarrow}&{2k-4} & {~=~}& {14 +14k } &{} \\ {\Rightarrow}&{-12k} & {~=~}& {18} &{} \\ {\Rightarrow}&{-2k} & {~=~}& {3} &{} \\ {\Rightarrow}&{k} & {~=~}& {- \frac{3}{2}} &{} \\ \end{array}$

(ii) If k = -3/2, then the y-coordinate can be calculated from the result in (4). We get:

$\begin{array}{ll} {}&{\rm{y-coordinate}} & {~=~}& {\frac{4k + 6}{1+k}} &{} \\ {}&{} & {~=~}& {\frac{4(-3/2) + 6}{1+(-3/2)}} &{} \\ {}&{} & {~=~}& {\frac{-6+6}{1+(-3/2)}} &{} \\ {}&{} & {~=~}& {0} &{} \\ \end{array}$

• Note that, the y-coordinate of C is also '0'.

(iii) If k = -3/2, then the z-coordinate can be calculated from the result in (4). We get:

$\begin{array}{ll} {}&{\rm{z-coordinate}} & {~=~}& {\frac{6k + 10}{1+k}} &{} \\ {}&{} & {~=~}& {\frac{6(-3/2) + 10}{1+(-3/2)}} &{} \\ {}&{} & {~=~}& {\frac{-9+10}{-(1/2)}} &{} \\ {}&{} & {~=~}& {\frac{1}{-(1/2)}} &{} \\ {}&{} & {~=~}& {-2} &{} \\ \end{array}$

• Note that, the z-coordinate of C is also '-2'.

(iv) So the coordinates of C satisfy the conditions in (4).

6. That means, the point C indeed divides AB internally or externally in the ratio k:1
Since k (= -3/2) is a -ve value, it is an external division.
7. Point C can divide AB (internally or externally) only if A, B and C is collinear.

Solved example 12.9
Find the coordinates of the centroid of the triangle PQR whose vertices are P(x₁,y₁,z₁), Q(x₂,y₂,z₂), R(x₃,y₃,z₃).
Solution:
1. Let A be the midpoint of QR. Then the coordinates of A will be:
$\left(\frac{x_2 + x_3}{2},~\frac{y_2 + y_3}{2},~\frac{z_2 + z_3}{2} \right)$
2. Let G be the centroid.
3. We have the coordinates of P and A.
• The centroid G will divide PA internally in the ratio 2:1
• So the coordinates of G will be:
$\left(\frac{m x_2 + n x_1}{m+n},~\frac{m y_2 + n y_1}{m+n},~\frac{m z_2 + n z_1}{m+n} \right)$
⇒ $\left(\frac{2 × \frac{x_2 + x_3}{2} + 1 × x_1}{2+1},~\frac{2 × \frac{y_2 + y_3}{2} + 1 × y_1}{2+1},~\frac{2 × \frac{z_2 + z_3}{2} + 1 × z_1}{2+1} \right)$
⇒ $\left(\frac{x_2 + x_3 + 1 × x_1}{2+1},~\frac{y_2 + y_3 + 1 × y_1}{2+1},~\frac{z_2 + z_3 + 1 × z_1}{2+1} \right)$
⇒ $\left(\frac{x_1 + x_2 + x_3}{3},~\frac{y_1 + y_2 + y_3}{3},~\frac{z_1 + z_2 + z_3}{3} \right)$

Solved example 12.10
Find the ratio in which the line segment joining the points A(4,8,10) and B(6,10,-8) is divided by the YZ-plane. Also find the coordinates of the point at which the division is done.
Solution:
1. Let the YZ-plane divide AB in the ratio k:1
2. Let the point of division be C.
3. We can write the coordinates of C as follows:
$\left(\frac{k x_2 + x_1}{1+k},~\frac{k y_2 + y_1}{1+k},~\frac{k z_2 + z_1}{1+k} \right)$
⇒ $\left(\frac{k × 6 + 4}{1+k},~\frac{k × 10 + 8}{1+k},~\frac{k × -8 + 10}{1+k} \right)$
⇒ $\left(\frac{6k + 4}{1+k},~\frac{10k + 8}{1+k},~\frac{-8k + 10}{1+k} \right)$
4. Consider any point on the YZ-plane. The x-coordinate of that point will be zero.
So the x-coordinate written in (3) is zero. We get:

$\begin{array}{ll} {}&{\frac{6k + 4}{1+k}} & {~=~}& {0} &{} \\ {\Rightarrow}&{6k + 4} & {~=~}& {0} &{} \\ {\Rightarrow}&{6k} & {~=~}& {-4} &{} \\ {\Rightarrow}&{3k} & {~=~}& {-2} &{} \\ {\Rightarrow}&{k} & {~=~}& {- \frac{2}{3}} &{} \\ \end{array}$

5. Thus we get the value of k.
• We have seen that, k:1 is just another form of m:n. Both represent the same ratio.
So we can write:

$\begin{array}{ll} {}&{k:1} & {~=~}& {\frac{k}{1}} &{} \\ {\Rightarrow}&{\frac{k}{1}} & {~=~}& {\frac{m}{n}} &{} \\ {\Rightarrow}&{\frac{-(2/3)}{1}} & {~=~}& {\frac{m}{n}} &{} \\ {\Rightarrow}&{-\frac{2}{3}} & {~=~}& {\frac{m}{n}} &{} \\ \end{array}$

• That means, the YZ-plane divides AB in the ratio m:n where m/n = -(2/3)

6. Now we can calculate the y-coordinate of C.
• Substituting the value of k in the expression for y-coordinate in (3), we get:

$\begin{array}{ll} {}&{\rm{y-coordinate}} & {~=~}& {\frac{10k + 8}{1+k}} &{} \\ {}&{} & {~=~}& {\frac{10(-2/3) + 8}{1+(-2/3)}} &{} \\ {}&{} & {~=~}& {\frac{-20+24}{3-2}} &{} \\ {}&{} & {~=~}& {4} &{} \\ \end{array}$

7. Finally we can calculate the z-coordinate of C.
• Substituting the value of k in the expression for z-coordinate in (3), we get:

$\begin{array}{ll} {}&{\rm{z-coordinate}} & {~=~}& {\frac{-8k + 10}{1+k}} &{} \\ {}&{} & {~=~}& {\frac{-8(-2/3) + 10}{1+(-2/3)}} &{} \\ {}&{} & {~=~}& {\frac{16+30}{3-2}} &{} \\ {}&{} & {~=~}& {46} &{} \\ \end{array}$

8. So the division is done at C(0,4,46)
• Fig.12.18 below shows the actual plot.

Fig.12.18

We can write 3 points:
(i) We see that:
• The YZ-plane does not divide AB
• But the YZ plane divides the extension of AB. So it is an external division. The division is done at C(0,4,46)
(ii) We have the coordinates of all three points:
A(4,8,10), B(6,10,-8) and C(0,4,46)
    ♦ Using the distance formula, BC = √2988
    ♦ Using the distance formula, AC = √1328
(iii) So we get:
$\frac{AC}{BC}~=~\frac{\sqrt{1328}}{\sqrt{2988}}~=~\sqrt{\frac{1328}{2988}}~=~\sqrt{\frac{4}{9}}~=~\frac{2}{3}$

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The link below gives a few more solved examples:

Exercise 12.3

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In the next section, we will see some miscellaneous examples.

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