Question

A point R with x-coordinate 4 lies on the line segment joining the points P(2, -3, 4) and Q(8, 0, 10). Find the coordinates of the point R.

Answer 1

Hint: Assume that the point ‘R’ is dividing the length of the line segment joining the coordinates P(2,-3,4) and Q(8,0,10) in ratio K:1. Now, use the section formula to find the other coordinates (y and z) of point R.

Complete step-by-step solution:
Let us suppose coordinates of point R be $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$.
We know the x-coordinate of ‘R’ is ‘4’ from the question. Hence,
${{x}_{1}}=4.............\left( i \right)$
Now, we have another two points P and Q given as P(2,-3,4) and Q(8,0,10), such that point R is lying on the line joining PQ.

Let us suppose point R is dividing the length PQ in the ratio K:1 as shown in the diagram.
Now, we can use the section formula to get coordinates of ‘R’ in terms of points P and Q and the given ratio K:1.
Section formula can be given as:
If any point C is dividing length AB in the ratio m:n with coordinates of A and B as \[\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\], then coordinates of C can be given by relation,
$C=\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\dfrac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)$
Hence, point R from the figure (2) can be given as,
$\begin{align}
  & R\left( 4,{{y}_{1}},{{z}_{1}} \right)=\left( \dfrac{8k+2}{k+1},\dfrac{0.k-1\times 3}{k+1},\dfrac{10k+4}{k+1} \right) \\
 & R\left( 4,{{y}_{1}},{{z}_{1}} \right)=\left( \dfrac{8k+2}{k+1},\dfrac{-3}{k+1},\dfrac{10k+4}{k+1} \right)..............\left( ii \right) \\
\end{align}$
Since, coordinates from the above relation should be equal. Hence, on equating x-coordinate, we get
$4=\dfrac{8k+2}{k+1}$
On cross multiplication, we get
 $\begin{align}
  & 4k\text{ }+\text{ }4\text{ }=\text{ }8K\text{ }+\text{ }2 \\
 & 2\text{ }=\text{ }4k \\
 & k=\dfrac{1}{2} \\
\end{align}$
Hence, from the equation (ii), we can get values of $\left( {{y}_{1}} \right)$ and $\left( {{z}_{1}} \right)$ as,
${{y}_{1}}=\dfrac{-3}{k+1}=\dfrac{-3}{\left( \dfrac{1}{2}+1 \right)}=\dfrac{-3\times 2}{3}=-2$
And ${{z}_{1}}$ can be given as,
${{z}_{1}}=\dfrac{10k+}{k+1}=\dfrac{10\times \dfrac{1}{2}+4}{\dfrac{1}{2}+1}=\dfrac{5+4}{\left( \dfrac{3}{2} \right)}=\dfrac{9\times 2}{3}=6$
Hence, the coordinates of point R can be given as R(4,-2,6).

Note: We can get equation of line PQ by relation $\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}$, where $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ are points on line.
Find the equation of the line and put x = 4 in that equation to get other coordinates of point R.
One can go wrong with the sectional formula. He or she may write the formula as
$\left( \dfrac{m{{x}_{1}}+n{{x}_{2}}}{m+n},\dfrac{m{{y}_{1}}+n{{y}_{2}}}{m+n},\dfrac{m{{z}_{1}}+n{{z}_{2}}}{m+n} \right)$
This is wrong. So, be careful with the place of terms to get the correct answer.