Question

If the points $A(2,0,0),B(4,p,q)$ and $C(1,5,2)$ are collinear,find the values of $p$ and $q$

Answer 1

Given that the points $A(2,0,0),B(4,p,q)$ 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

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

But the coordinates of the point $B$ are $(4,p,q)$, so we have,

$\frac{k+2}{k+1}=4,\frac{5k}{k+1}=p,\frac{2k}{k+1}=q$

$\Rightarrow k+2=4k+4,\frac{5k}{k+1}=p,\frac{2k}{k+1}=q$

$\Rightarrow k-4k=4-2,\frac{5k}{k+1}=p,\frac{2k}{k+1}=q$

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

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

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

$\Rightarrow k=\frac{-2}{3},p=\frac{\frac{-10}{3}}{\frac{-2+3}{3}},q=\frac{\frac{-4}{3}}{\frac{-2+3}{3}}$

$\Rightarrow k=\frac{-2}{3},p=\frac{\frac{-10}{3}}{\frac{1}{3}},q=\frac{\frac{-4}{3}}{\frac{-1}{3}}$

$\therefore k=\frac{-2}{3},p=-10,q=4$