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 $P,Q,R$ and $S$.

• A. $P=\frac{8}{3},Q=\frac{7}{3},R=\frac{-4}{3},S=\frac{-5}{3}$
• B. $P=\frac{7}{3},Q=\frac{7}{3},R=\frac{-5}{3},S=\frac{-4}{3}$
• C. $P=\frac{7}{3},Q=\frac{8}{3},R=\frac{-4}{3},S=\frac{-5}{3}$
• D. $P=\frac{8}{3},Q=\frac{7}{3},R=\frac{-5}{3},S=\frac{-4}{3}$

Answer 1

The correct option is C $P=\frac{7}{3},Q=\frac{8}{3},R=\frac{-4}{3},S=\frac{-5}{3}$

from the given number line, we can observe that the points $U,T$ are $-1-(-2)=1$ unit distance apart.

Given that the points are such that $TR=RS=SU$.

Therefore, $TR=RS=SU=\frac{1}{3}$.

The point $S$ is at $-2+\frac{1}{3}=-\frac{5}{3}$

The point $R$ is at $-2+\frac{1}{3}+\frac{1}{3}=-\frac{4}{3}$

from the given number line, we can observe that the points $A,B$$are$3-2=1$ unit distance apart.

Given that the points are such that $AP=PQ=QB$.

Therefore, $AP=PQ=QB=\frac{1}{3}$.

The point $P$ is at $2+\frac{1}{3}=\frac{7}{3}$

The point $Q$ is at $2+\frac{1}{3}+\frac{1}{3}=\frac{8}{3}$