Question

The points with the coordinates $ (2a,3a),\,(3b,2b)\,{\text{and}}\,(c,c) $ are collinear:
A.For no value of $ a,\,b,\,c $
B.For all value of $ a,\,b,\,c $
C.If $ a,\,\dfrac{c}{5} $ are in HP
D.If $ a,\,\dfrac{{2c}}{5},\,b $ are in HP

Answer 1

Hint: Here before solving this question we need to know the following formula: -
Using determinant points are collinear if
 $ \left| {\begin{array}{*{20}{c}}
  {{x_1}}&{{y_1}}&1 \\
  {{x_2}}&{{y_2}}&1 \\
  {{x_3}}&{{y_3}}&1
\end{array}} \right| = 0\, $
Where,
 $ ({x_1},{y_1}),({x_2},{y_2})\,{\text{and}}\,({x_3},{y_3}) $ are the coordinates

Complete step-by-step answer:
According to this question we have,
 $ \begin{gathered}
  ({x_1},{y_1}) = (2a,3a) \\
  ({x_2},{y_2}) = (3b,2b)\,{\text{and}}\, \\
  ({x_3},{y_3}) = (c,c) \\
\end{gathered} $
Substitute this values on equation (1)
 $ \begin{gathered}
  \left| {\begin{array}{*{20}{c}}
  {2a}&{3a}&1 \\
  {3b}&{2b}&1 \\
  c&c&1
\end{array}} \right| = 0 \\
  {\text{Applying}}\,{R_2} \to {R_2} - {R_1}\,{\text{and}}\,{R_3} \to {R_3} - {R_1} \\
  \left| {\begin{array}{*{20}{c}}
  {2a}&{3a}&1 \\
  {3b - 2a}&{2b - 3a}&0 \\
  {c - 2a}&{c - 3a}&0
\end{array}} \right| = 0 \\
\end{gathered} $
Expanding along column 3, we get
 $ \begin{gathered}
   = \left| {\begin{array}{*{20}{c}}
  {3b - 2a}&{2b - 3a} \\
  {c - 2a}&{c - 3a}
\end{array}} \right| = 0 \\
  \left( {3b - 2a} \right)\left( {c - 3a} \right) - \left( {2b - 3a} \right)\left( {c - 2a} \right) = 0 \\
  \left( {3bc - 9ab - 2ac + 6{a^2}} \right) - \left( {2bc - 4ab - 3ac + 6{a^2}} \right) = 0 \\
  3bc - 9ab - 2ac + 6{a^2} - 2bc + 4ab + 3ac - 6{a^2} = 0 \\
   - 5ab + bc + ac = 0 \\
  bc + ac = 5ab \\
  c(a + b) = 5ab \\
  \dfrac{c}{5} = \dfrac{{ab}}{{a + b}} \\
  \dfrac{{2c}}{5} = \dfrac{{2ab}}{{a + b}} \\
\end{gathered} $
Thus, $ a,\dfrac{{2c}}{5},b $ are inHP
So, the correct answer is “Option D”.

Note: Determinant properties are used before expanding the determinant. Also, two numbers a and b are in HP can be shown as $ \dfrac{1}{a},\dfrac{1}{b} $ and adding this we get $ \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{{a + b}}{{ab}} $