# Show that the three points A (2, 3, 4),

Question:

Show that the three points A (2, 3, 4), B (–1, 2, – 3) and C (– 4, 1, – 10) are collinear and find the ratio in which C divides AB.

Solution:

Given three points are $A(2,3,4), B(-1,2,-3)$ and $C(-4,1,-10)$ To find collinear points,

$\mathrm{AB}=\sqrt{(2+1)^{2}+(3-2)^{2}+(4+3)^{2}}$

$=\sqrt{9+1+49}$

$=\sqrt{59}$

Now consider,

$B C=\sqrt{(-1+4)^{2}+(2-1)^{2}+(-3+10)^{2}}$

$=\sqrt{9+1+49}$

$=\sqrt{59}$

Again we have,

$\mathrm{AC}=\sqrt{(2+4)^{2}+(3-1)^{2}+(4+10)^{2}}$

$=\sqrt{36+4+196}$

$=\sqrt{236}$

$=2 \sqrt{59}$

$\Rightarrow \mathrm{AB}+\mathrm{BC}=\mathrm{AC} ;$ Points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are collinear.

$A C: B C=2 \sqrt{59}: \sqrt{59}=2: 1$

From the lengths of $A B, B C$ and $A C$ we can say that $C$ divides $A B$ in the ratio $2: 1$ externally.

Again we have,

$\mathrm{AC}=\sqrt{(2+4)^{2}+(3-1)^{2}+(4+10)^{2}}$

$=\sqrt{36+4+196}$

$=\sqrt{2} 36$

$=2 \sqrt{59}$

$\Rightarrow \mathrm{AB}+\mathrm{BC}=\mathrm{AC} ;$ Points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are collinear.

$A C: B C=2 \sqrt{5} 9: \sqrt{59}=2: 1$

From the lengths of $A B, B C$ and $A C$ we can say that $C$ divides $A B$ in the ratio $2: 1$ externally.