Question

If the points $A(1,0,-2),B(2,p,q)$ and $C(-1,5,2)$ are collinear,find the values of $p$ and $q$

Answer 1

Given that the points $A(1,0,-2),B(2,p,q)$ and $C(-1,5,2)$ 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

$(\frac{-k+1}{k+1},\frac{5k+0}{k+1},\frac{2k-2}{k+1})$

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

$\frac{-k+1}{k+1}=2,\frac{5k}{k+1}=p,\frac{2k-2}{k+1}=q$

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

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

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

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

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

$\Rightarrow k=\frac{-1}{3},\frac{\frac{-5}{3}}{\frac{2}{3}}=p,\frac{\frac{-2-6}{3}}{\frac{2}{3}}=q$

$\Rightarrow k=\frac{-1}{3},p=\frac{\frac{-5}{3}}{\frac{2}{3}},q=\frac{\frac{-8}{3}}{\frac{2}{3}}$

$\Rightarrow k=\frac{-1}{3},p=\frac{-5}{2},q=\frac{\frac{-8}{3}}{\frac{2}{3}}$

$\Rightarrow k=\frac{-1}{3},p=\frac{-5}{2},q=\frac{-8}{2}$

$\therefore k=\frac{-1}{3},p=\frac{-5}{2},q=-4$