Question

The points $P(a,b+c),Q(b,c+a)$ and $R(c,a+b)$ are such that $PQ=QR$ if

• A. $a,b,c$ are in A.P.
• B. $a,b,c$ are in G.P.
• C. $a,b,c$ are in H.P.
• D. None of these

Answer 1

The correct option is A

$a,b,c$ are in A.P.

Explanation for the correct option:

Step 1: Compute the required distances

In the question, Three points $P(a,b+c),Q(b,c+a)$ and $R(c,a+b)$ are given.

Compute the distance between $P$ and $Q$.

$$\begin{gathered} PQ=\sqrt{{(b-a)}^2+{(c+a-b-c)}^2} \\ \Rightarrow PQ=\sqrt{{(b-a)}^2+{(a-b)}^2} \\ \Rightarrow PQ=\sqrt{{(a-b)}^2+{(a-b)}^2} \\ \Rightarrow PQ=\sqrt{{2(a-b)}^2} \end{gathered}$$

Compute the distance between $Q$ and $R$.

$$\begin{gathered} QR=\sqrt{{(c-b)}^2+{(a+b-c-a)}^2} \\ \Rightarrow QR=\sqrt{{(b-c)}^2+{(b-c)}^2} \\ \Rightarrow QR=\sqrt{{2(b-c)}^2} \end{gathered}$$

Step 2: Find the relation between distances

Since, it is given that $PQ=QR$.

So, $\sqrt{2{(a-b)}^2}=\sqrt{2{(b-c)}^2}$.

$$\begin{gathered} \left(a-b\right)=\left(b-c\right) \\ \Rightarrow a+c=2b \end{gathered}$$

This implies that, $a,b$ and $c$ are in A.P.

Therefore, If the points $P(a,b+c),Q(b,c+a)$ and $R(c,a+b)$ are such that $PQ=QR$, then $a,b$ and $c$ are in A.P.

Hence, option $A$ is the correct answer.