Question

Show that the points $A(3,2,-4),B(5,4,-6)$ and $C(9,8,-10)$ are collinear, find the ratio in which $B$ divides $\overline{AC}$.

Answer 1

Given that

Point A(3,2,-4), B(5,4,-6) & C(9,8,-10) are collinear

B must divide line segment AC in some ratio externally and internally

We know that

Co-ordinate of point $A(x,y,z)$ that divides line segment joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in ratio $m:n$ is

$(x,y,z)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Let point $B(5,4,-6)$ divide line segment $A(3,2,-4)$ and $C(9,8,-10)$ in the ratio $k:1$

$B(5,4,-6)=(\frac{9k+3}{k+1},\frac{8k+2}{k+1},\frac{-10k-4}{k+1})$

Comparing x-coordinate of B

$5=\frac{9k+3}{k+1}$

$5k+5=9k+3$

$9k-5k=5-3$

$4k=2$

$k=\frac{1}{2}$

So, $k:1=1:2$

Thus, point B divides AC in the ratio 1:2