Question

If the points $A(-1,-2,1),B(-2,p,q)$ and $C(1,-2,-1)$ are collinear,find the values of $p$ and $q$.

Answer 1

Given that the points $A(-1,-2,1),B(-2,p,q)$ and $C(1,-2,-1)$ are collinear.Let the point $B$ divide the line segment $AC$ in the ratio $k:1$, then by section formula the coordinates of the point $B$ are

$(\frac{k-1}{k+1},\frac{-2k-2}{k+1},\frac{-k+1}{k+1})$

But the coordinates of the point $B$ are $(-2,p,q)$, so we have,

$\frac{k-1}{k+1}=-2,\frac{-2k-2}{k+1}=p,\frac{-k+1}{k+1}=q$

$\Rightarrow k-1=-2k-2,\frac{-2(k+1)}{k+1}=p,\frac{-k+1}{k+1}=q$

$\Rightarrow k+2k=-2+1,p=-2,\frac{-k+1}{k+1}=q$

$\Rightarrow 3k=-1,p=-2,\frac{-k+1}{k+1}=q$

$\Rightarrow k=\frac{-1}{3},p=-2,q=\frac{-k+1}{k+1}$

$\Rightarrow k=\frac{-1}{3},p=-2,q=\frac{-\frac{-1}{3}+1}{\frac{-1}{3}+1}$

$\Rightarrow k=\frac{-1}{3},p=-2,q=\frac{\frac{1+3}{3}}{\frac{-1+3}{3}}$

$\Rightarrow k=\frac{-1}{3},p=-2,q=\frac{\frac{4}{3}}{\frac{2}{3}}$

$\Rightarrow k=\frac{-1}{3},p=-2,q=\frac{4}{2}$

$\therefore k=\frac{-1}{3},p=-2,q=2$