Prove that the points (0, – 1, – 7), (2, 1, – 9) and (6, 5, – 13) are collinear. Find the ratio in which the first point divides the join of the other two.

Home » Prove that the points (0, – 1, – 7), (2, 1, – 9) and (6, 5, – 13) are collinear. Find the ratio in which the first point divides the join of the other two.

CBSE | Class 11 | Introduction to Three Dimensional Geometry | Maths | NCERT Exemplar

Prove that the points (0, – 1, – 7), (2, 1, – 9) and (6, 5, – 13) are collinear. Find the ratio in which the first point divides the join of the other two.

Solution:

It is given that the three points $A(0,-1,-7), B(2,1,-9)$ and $C(6,5,-13)$ are collinear
So it can be written as
$\begin{array}{l} A B=\sqrt{(2-0)^{2}+(1-(-1))^{2}+((-9)-(-7))^{2}}=\sqrt{4+4+4}=2 \sqrt{3} \\ B C=\sqrt{(6-2)^{2}+(5-1)^{2}+((-13)-(-9))^{2}}=\sqrt{16+16+16}=4 \sqrt{3} \\ A C=\sqrt{(6-0)^{2}+(5-(-1))^{2}+((-13)-(-7))^{2}}=\sqrt{36+36+36}=6 \sqrt{3} \\ \Rightarrow A B+B C=A C \end{array}$
As points $A, B$ and $C$ are collinear.
$\mathrm{AB}: \mathrm{AC}=2 \sqrt{3}: 6 \sqrt{3}=1: 3$
As a result, from the lengths of $A B, B C$ and $A C$ we can say that the first point divides the join of the other two in the ratio $1: 3$ externally.