If the points \[(a,0),{\text{ }}(0,b)\] and \[(3,2)\] are collinear, prove that, \[\dfrac{2}{b} + \dfrac{3}{a} = 1\].

Hint: In this question, we have to prove the particular for the collinear points.
In the case of geometry, if three given points A, B, and C are collinear then the area of the triangle formed by three collinear is zero.
First we need to find out the area of the triangle formed by three points then equating it to zero we will get the required result.

Formula used: The area of the triangle formed by three points A\[\left( {{x_1},{y_1}} \right)\], B\[\left( {{x_2},{y_2}} \right)\], C\[\left( {{x_3},{y_3}} \right)\] is \[\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\]

Complete step-by-step answer:
It is given that the points \[(a,0),{\text{ }}(0,b)\] and \[(3,2)\] are collinear.
We need to prove that, \[\dfrac{2}{b} + \dfrac{3}{a} = 1\].
Let us consider the points are A =\[(a,0)\], B =\[(0,b)\] and C= \[(3,2)\] and these are collinear.
The area of the triangle formed by three points A\[(a,0)\], B\[(0,b)\] and C\[(3,2)\] is
$\Rightarrow$\[\dfrac{1}{2}\left[ {a\left( {b - 2} \right) + 0\left( {2 - 0} \right) + 3\left( {0 - b} \right)} \right] = \dfrac{1}{2}\left( {ab - 2a - 3b} \right)\]
We know that, if three given points A, B, and C are collinear then the area of the triangle formed by three collinear is zero.
Thus we get, \[\dfrac{1}{2}\left( {ab - 2a - 3b} \right) = 0\]
By cross multiplication we get,
$\Rightarrow$\[\left( {ab - 2a - 3b} \right) = 0\]
$\Rightarrow$\[2a + 3b = ab\]
Dividing both sides by \[ab\] we get,
$\Rightarrow$\[\dfrac{{2a + 3b}}{{ab}} = \dfrac{{ab}}{{ab}}\]
$\Rightarrow$\[\dfrac{2}{b} + \dfrac{3}{a} = 1\]

Hence we get if the points \[(a,0),{\text{ }}(0,b)\] and \[(3,2)\] are collinear, then \[\dfrac{2}{b} + \dfrac{3}{a} = 1\](proved).

Note: A set of points are said to be collinear if they all lie on a single line. Slope of the collinear points is the same. The slope of perpendicular lines is equal to -1.