Question

Find the correct option
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

Hint: To find relation between a,b and c. first we will calculate lengths of \[PQ\] and \[QR\] in terms of the variables a, b, c and then we will equate both lengths and simplify the equation to get relation between a, b and c.

Formulae Used: The following formulas have been used to solve this problem:
1. If \[d\] is the distance between two points \[({x_1},{y_1})\] and \[({x_2},{y_2})\] , then \[d = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}} \] .
2. If a, b and c are in A.P. , then \[a + c = 2b\]
3. \[{(p - q)^2} = {p^2} - 2pq + {q^2}\]

Complete step-by-step solution:
We have been given three points as \[P(a,b + c)\] , \[Q(b,c + a)\] and \[R(c,a + b)\] .
First, we will calculate the distance \[PQ\] using distance formula \[d = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}} \] .
Assigning the given values to the variables in the formula, we have
\[\begin{array}{l}{x_1} = a\\{y_1} = b + c\end{array}\] and \[\begin{array}{l}{x_2} = b\\{y_2} = c + a\end{array}\]
Substituting the values
\[PQ = \sqrt {{{(a - b)}^2} + {{\{ (b + c) - (c + a)\} }^2}} \]
\[ = \sqrt {{{(a - b)}^2} + {{(b + c - c - a)}^2}} \]
\[ = \sqrt {{{(a - b)}^2} + {{(b - a)}^2}} \]
Further solving the above
\[PQ = \sqrt {{{(a - b)}^2} + {{(a - b)}^2}} \] [Since, \[{(a - b)^2} = {(b - a)^2}\] ]
\[ = \sqrt {2{{(a - b)}^2}} \]
\[ = \sqrt 2 (a - b)\]
Similarly, we will calculate the distance \[QR\] .
Assigning the given values to the variables in the formula, we have
\[\begin{array}{l}{x_1} = b\\{y_1} = c + a\end{array}\] and \[\begin{array}{l}{x_2} = c\\{y_2} = a + b\end{array}\]
Substituting the values
\[QR = \sqrt {{{(b - c)}^2} + {{\{ (c + a) - (a + b)\} }^2}} \]
\[ = \sqrt {{{(b - c)}^2} + {{(c + a - a - b)}^2}} \]
\[ = \sqrt {{{(b - c)}^2} + {{(c - b)}^2}} \]
Further solving the above
\[QR = \sqrt {{{(b - c)}^2} + {{(b - c)}^2}} \] [Since, \[{(b - c)^2} = {(c - b)^2}\] ]
\[ = \sqrt {2{{(b - c)}^2}} \]
\[ = \sqrt 2 (b - c)\]
Now, we will equate \[PQ\] to \[RQ\] as per given data.
\[PQ = QR\]
\[ \Rightarrow \sqrt 2 (a - b) = \sqrt 2 (b - c)\]
\[ \Rightarrow a - b = b - c\]
\[ \Rightarrow a + c = 2b\] , which implies a, b and c are in A.P.
Hence, option A. is the correct answer.

Note: By solving the equation formed by the given data, if it is found that \[{b^2}\] is equal to \[ac\] , then a, b, c will be in G.P., but, if \[b\] will be equal to \[\dfrac{{2ac}}{{a + c}}\] , then a, b, c will be in H.P.