Question

If the points $A\left( 1,0,-6 \right),B\left( -3,p,q \right)\text{ and }C\left( -5,9,6 \right)$ are collinear, find the values of p and q.

Answer 1

Hint: In this question, we are given three points that are collinear and we have to find the value of p and q which are among the given points. For this, we will use the approach that, since all points lie on the same line so one point will divide the line joining the other two points in the same ratio k:1. Then using section formula we will be able to find the required value. For any two points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ whose line is being divided by the point $\left( x,y,z \right)$ in the ratio the section formula is given as,
$\left( x,y,z \right)=\left( \dfrac{{{x}_{2}}k+{{x}_{1}}}{k+1},\dfrac{{{y}_{2}}k+{{y}_{1}}}{k+1},\dfrac{{{z}_{2}}k+{{z}_{1}}}{k+1} \right)$.

Complete step-by-step answer:
Here we are given the point as $A\left( 1,0,-6 \right),B\left( -3,p,q \right)\text{ and }C\left( -5,9,6 \right)$. We are given that all these points are collinear, therefore they lie on the same line. Hence we can suppose that point B cut the lines joining point A and C in some ratio k:1.

We know that, if a point $\left( x,y,z \right)$ divides a line joining the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ in the ratio k:1, then section formula is given by,
$\left( x,y,z \right)=\left( \dfrac{{{x}_{2}}k+{{x}_{1}}}{k+1},\dfrac{{{y}_{2}}k+{{y}_{1}}}{k+1},\dfrac{{{z}_{2}}k+{{z}_{1}}}{k+1} \right)$.
Here B divides the line joining A and C in the ratio k:1, therefore, values become
\[\begin{align}
  & x=-3,y=p,z=q \\
 & {{x}_{1}}=1,{{y}_{1}}=0,{{z}_{1}}=-6 \\
 & {{x}_{2}}=-5,{{y}_{2}}=9,{{z}_{2}}=6 \\
\end{align}\]
Putting in the values we get:
$\left( -3,p,q \right)=\left( \dfrac{-5k+1}{k+1},\dfrac{9k+0}{k+1},\dfrac{6k-6}{k+1} \right)$.
By comparing we get:
\[\begin{align}
  & -3=\dfrac{-5k+1}{k+1}\cdots \cdots \cdots \left( 1 \right) \\
 & p=\dfrac{9k}{k+1}\cdots \cdots \cdots \left( 2 \right) \\
 & q=\dfrac{6k-6}{k+1}\cdots \cdots \cdots \left( 3 \right) \\
\end{align}\]
Let us solve them one by one to find our values, first let us find the value of k using equation (1) we have,
$-3=\dfrac{-5k+1}{k+1}$.
Cross multiplying we get:
$\begin{align}
  & -3\left( k+1 \right)=-5k+1 \\
 & \Rightarrow -3k-3=-5k+1 \\
\end{align}$
Taking variables on one side and constant on the other side we get:
$\begin{align}
  & \Rightarrow -3k+5k=1+4 \\
 & \Rightarrow 2k=4 \\
\end{align}$
Dividing both sides by 2, we get:
$k=2$.
Now let us use the value of k directly in equation (2) and (3) to get the value of p and q.
From equation (2) we have $p=\dfrac{9k}{k+1}$.
Putting k = 2, we have, $p=\dfrac{9\times 2}{2+1}=\dfrac{18}{3}=6$.
Hence the value of p is 6.
From equation (3) we have, $q=\dfrac{6k-6}{k+1}$.
Putting k = 2, we have $q=\dfrac{6\times 2-6}{2+1}=\dfrac{12-6}{3}=\dfrac{6}{3}=2$.
Hence the value of q is 2.
Therefore, the required value of p and q are 6, 2 respectively.

Note: Students should note that, if the value of k is found as negative, then this means that point B lies outside the line joining points A and C. So we would need a new diagram for that. Make sure that formula is used as such, if points are exchanged, then the ratio also should get exchanged.