Question

If the points $A(0,-1,-3),B(1,p,q)$ and $C(-2,6,3)$ are collinear,find the values of $p$ and $q$

Answer 1

Given that the points $A(0,-1,-3),B(1,p,q)$ and $C(-2,6,3)$ 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

$(\frac{-2k+0}{k+1},\frac{6k-1}{k+1},\frac{3k-3}{k+1})$

But the coordinates of the point $B$ are $(1,p,q)$, so we have,

$\frac{-2k}{k+1}=1,\frac{6k-1}{k+1}=p,\frac{3k-3}{k+1}=q$

$\Rightarrow -2k=k+1,\frac{6k-1}{k+1}=p,\frac{3k-3}{k+1}=q$

$\Rightarrow -2k-k=1,\frac{6k-1}{k+1}=p,\frac{3k-3}{k+1}=q$

$\Rightarrow -3k=1,\frac{6k-1}{k+1}=p,\frac{3k-3}{k+1}=q$

$\Rightarrow k=\frac{-1}{3},\frac{6k-1}{k+1}=p,\frac{3k-3}{k+1}=q$

$\Rightarrow k=\frac{-1}{3},\frac{6\times \frac{-1}{3}-1}{\frac{-1}{3}+1}=p,\frac{3\times \frac{-1}{3}-3}{\frac{-1}{3}+1}=q$

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

$\Rightarrow k=\frac{-1}{3},p=\frac{-9}{2},q=\frac{-12\times 3}{2}$

$\therefore k=\frac{-1}{3},p=\frac{-9}{2},q=-18$