Question

The points $P$, $Q$, $R$, $S$, $T$, $U$, $A$ and $B$ on the number line are such that $TR=RS=SU$ and $AP=PQ=QB$. Name the rational numbers represented by $P$, $Q$, $R$, $S$.

Answer 1

Hint: From the given diagram we have the values of $A$, $B$, $U$, $T$. From this value we will find the values of f$AB$ and $UT$. From the given diagram we can write $AB$ as $AP+PQ+QB$, we have $AP=PQ=QB$ . From this we can obtain any one of the values of $AP$, $PQ$, $QB$. Similarly, we will calculate the values of $TR$, $RS$, $SU$. To find the rational number of $P$ we will add $AP$ to $A$ and for finding a rational number of $Q$ we will add $A$, $AP$,$PQ$. Similarly, we will do for finding the rational numbers of $R$, $S$.

Complete step-by-step answer:
Given that,
Number line is

$TR=RS=SU$ and $AP=PQ=QB$
From the given number line we can write that $A=2$, $B=3$, $T=-1$, $U=-2$.
We can write the value of $AB$ from given number line as given below
$AB=AP+PQ+QB$
The value of $AB$ is $3-2=1$.
We have given that $AP=PQ=QB$
$\begin{align}
  & \Rightarrow AB=AP+PQ+QB \\
 & \Rightarrow 1=AP+AP+AP \\
 & \Rightarrow 1=3AP \\
 & \Rightarrow AP=\dfrac{1}{3} \\
\end{align}$
$\therefore AP=PQ=QB=\dfrac{1}{3}$.
From the line diagram the rational number of $P$ can be calculated as
$\begin{align}
  & P=A+AP \\
 & \Rightarrow P=2+\dfrac{1}{3} \\
 & \Rightarrow P=\dfrac{2\times 3+1}{3} \\
 & \Rightarrow P=\dfrac{7}{3} \\
\end{align}$
Again, the rational number of $Q$ can be calculated as
$\begin{align}
  & Q=A+AP+PQ \\
 & \Rightarrow Q=2+\dfrac{1}{3}+\dfrac{1}{3} \\
 & \Rightarrow Q=\dfrac{2\times 3+1+1}{3} \\
 & \Rightarrow Q=\dfrac{8}{3} \\
\end{align}$
From the number line we can write that $TU=TR+RS+SU$
Now the value of $TU$ is given by $-1-\left( -2 \right)=-1+2=1$
We have given that $TR=RS=SU$
$\begin{align}
  & \Rightarrow TU=TR+RS+SU \\
 & \Rightarrow 1=TR+TR+TR \\
 & \Rightarrow 1=3TR \\
 & \Rightarrow TR=\dfrac{1}{3} \\
\end{align}$
$\therefore TR=RS=SU=\dfrac{1}{3}$.
From the line diagram the rational number of $R$ can be calculated as
$\begin{align}
  & R=T-RT \\
 & \Rightarrow R=-1-\dfrac{1}{3} \\
 & \Rightarrow R=\dfrac{-1\times 3-1}{3} \\
 & \Rightarrow R=\dfrac{-4}{3} \\
\end{align}$
Again, the rational number of $S$ can be calculated as
$\begin{align}
  & S=T-RT-SR \\
 & \Rightarrow S=-1-\dfrac{1}{3}-\dfrac{1}{3} \\
 & \Rightarrow S=\dfrac{-1\times 3-1-1}{3} \\
 & \Rightarrow S=\dfrac{-5}{3} \\
\end{align}$

Note: While finding the rational number of $P$, $Q$ we added the values of $AP,PQ$ to the value of $A$, but while calculating the rational numbers of $P$, $S$ we have subtracted the values of $RT,SR$ because we need to go from right to left while moving from zero to negative values in number line but the values $RT,SR$ are indicates the left to right so we have subtracted them from $T$.