Question

Prove that the points \[\left( {a,0} \right),\left( {0,b} \right){\text{ and }}\left( {1,1} \right)\] are collinear if, \[\dfrac{1}{a} + \dfrac{1}{b} = 1\].

Answer 1

Hint: First of all, find the area of the triangle of the given points. The given points are collinear if the area formed by the three points is zero. Use the given condition to prove the collinearity of the given points.

Complete step-by-step answer:
Let the given points be \[{\text{A }}\left( {a,0} \right)\], \[{\text{B }}\left( {0,b} \right)\] and \[{\text{C }}\left( {1,1} \right)\]
Given \[\dfrac{1}{a} + \dfrac{1}{b} = 1\]
i.e., \[\dfrac{1}{a} + \dfrac{1}{b} - 1 = 0...........................................\left( 1 \right)\]
If the above points are collinear, they will lie on the same line i.e., they will not form a triangle.
Therefore, area of \[\Delta ABC = 0\]
We know that the area of the triangle formed by the points A \[\left( {{x_1},{y_1}} \right)\], B \[\left( {{x_2},{y_2}} \right)\] and C \[\left( {{x_3},{y_3}} \right)\] is given by \[\Delta = \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\]
So, area of \[\Delta ABC\] is given by
\[
   \Rightarrow \Delta = \dfrac{1}{2}\left[ {a\left( {b - 1} \right) + 0\left( {1 - 0} \right) + 1\left( {0 - b} \right)} \right] \\
   \Rightarrow \Delta = \dfrac{1}{2}\left[ {a\left( {b - 1} \right) + 0\left( 1 \right) + 1\left( { - b} \right)} \right] \\
   \Rightarrow \Delta = \dfrac{1}{2}\left[ {ab - a - b} \right] \\
\]
Multiplying and dividing with \[ab\]on left-hand side, we get
\[
   \Rightarrow \Delta = \dfrac{1}{2}\left( {ab} \right)\left[ {\dfrac{{ab - a - b}}{{ab}}} \right] \\
   \Rightarrow \Delta = \dfrac{{ab}}{2}\left[ {1 - \dfrac{a}{{ab}} - \dfrac{b}{{ab}}} \right] \\
   \Rightarrow \Delta = \dfrac{{ab}}{2}\left[ {1 - \dfrac{1}{b} - \dfrac{1}{a}} \right] \\
   \Rightarrow \Delta = \dfrac{{ - ab}}{2}\left[ {\dfrac{1}{a} + \dfrac{1}{b} - 1} \right] \\
   \Rightarrow \Delta = \dfrac{{ - ab}}{2}\left( 0 \right){\text{ }}\left[ {\because {\text{using }}\left( 1 \right)} \right] \\
  \therefore \Delta = 0 \\
\]
Hence the points \[\left( {a,0} \right),\left( {0,b} \right){\text{ and }}\left( {1,1} \right)\] are collinear.

Note: We can prove the collinearity of the three by another method i.e., if the sum of the lengths of any two-line segments among AB, BC, and CA is equal to the length of the remaining line segment, then the points are said to be in collinear.