Question

If the points $A(0,1,-1),B(3,p,q)$ and $C(0,4,1)$ are collinear,find the values of $p$ and $q$

Answer 1

Given that the points $A(0,1,-1),B(3,p,q)$ and $C(1,4,1)$ 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

$(\frac{k+0}{k+1},\frac{4k+1}{k+1},\frac{k-2}{k+1})$

But the coordinates of the point $B$ are $(3,p,q)$, so we have,

$\frac{k+0}{k+1}=3,\frac{4k+1}{k+1}=p,\frac{k-2}{k+1}=q$

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

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

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

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

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

$\Rightarrow k=\frac{-3}{2},\frac{-6+1}{\frac{-3+2}{2}}=p,\frac{\frac{-3-4}{2}}{\frac{-3+2}{2}}=q$

$\Rightarrow k=\frac{-3}{2},\frac{-5}{\frac{-1}{2}}=p,\frac{\frac{-7}{2}}{\frac{-1}{2}}=q$

$\therefore k=\frac{-3}{2},p=10,q=7$