If the points\[(a,0)\], \[(0,b)\] and \[(3,2)\] are collinear, prove that\[\dfrac{3}{a} + \dfrac{2}{b} = 1\].
Answer
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Hint: Here in this question we are given three points which are parallel to each other. We know that, when the coordinate points are parallel or collinear to each other, the slopes formed by them are equal. By using this property of collinear points, we can proceed further to prove that is required here.
Formula used: If \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] are the two points in a Cartesian plane \[X - Y\], then the slope formed by two points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is given by \[\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\].
Complete step-by-step answer:
We have three coordinate points\[(a,0)\], \[(0,b)\] and \[(3,2)\] which are collinear. Now we will find any two slopes formed between three collinear points. We, know that slope formed by points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is given by \[\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]. Using this formula we can find the slope between two points.
Slope formed by points \[(a,0)\] and \[(3,2)\] is given as\[\dfrac{{2 - 0}}{{3 - a}} = \dfrac{2}{{3 - a}}\].
Same way, slope formed by points \[(0,b)\] and \[(3,2)\] is given as \[\dfrac{{2 - b}}{{3 - 0}} = \dfrac{{2 - b}}{3}\].
Now, as we know that when the coordinate points are parallel or collinear to each other, the slopes forms by them are equal, we move ahead as,
\[
\dfrac{2}{{3 - a}} = \dfrac{{2 - b}}{3} \\
\Rightarrow 2 \times 3 = \left( {2 - b} \right)\left( {3 - a} \right) \\
\Rightarrow 6 = 6 - 2a - 3b + ab \\
\Rightarrow 2a + 3b = ab \\
\]
On dividing both LHS and RHS by \[ab\] we get,
\[
\Rightarrow \dfrac{{2a}}{{ab}} + \dfrac{{3b}}{{ab}} = \dfrac{{ab}}{{ab}} \\
\Rightarrow \dfrac{2}{b} + \dfrac{3}{a} = 1 \\
\]
Hence we have proved what was required.
Note: We can also prove the given question by area of triangle method, as three points are given. As we know that when three points with which we have a triangle, are collinear, then the area formed by that triangle will be zero, that is no such triangle can be formed. We can use the area of triangle formula and apply collinearity condition there to do the proof.
Formula used: If \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] are the two points in a Cartesian plane \[X - Y\], then the slope formed by two points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is given by \[\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\].
Complete step-by-step answer:
We have three coordinate points\[(a,0)\], \[(0,b)\] and \[(3,2)\] which are collinear. Now we will find any two slopes formed between three collinear points. We, know that slope formed by points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is given by \[\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]. Using this formula we can find the slope between two points.
Slope formed by points \[(a,0)\] and \[(3,2)\] is given as\[\dfrac{{2 - 0}}{{3 - a}} = \dfrac{2}{{3 - a}}\].
Same way, slope formed by points \[(0,b)\] and \[(3,2)\] is given as \[\dfrac{{2 - b}}{{3 - 0}} = \dfrac{{2 - b}}{3}\].
Now, as we know that when the coordinate points are parallel or collinear to each other, the slopes forms by them are equal, we move ahead as,
\[
\dfrac{2}{{3 - a}} = \dfrac{{2 - b}}{3} \\
\Rightarrow 2 \times 3 = \left( {2 - b} \right)\left( {3 - a} \right) \\
\Rightarrow 6 = 6 - 2a - 3b + ab \\
\Rightarrow 2a + 3b = ab \\
\]
On dividing both LHS and RHS by \[ab\] we get,
\[
\Rightarrow \dfrac{{2a}}{{ab}} + \dfrac{{3b}}{{ab}} = \dfrac{{ab}}{{ab}} \\
\Rightarrow \dfrac{2}{b} + \dfrac{3}{a} = 1 \\
\]
Hence we have proved what was required.
Note: We can also prove the given question by area of triangle method, as three points are given. As we know that when three points with which we have a triangle, are collinear, then the area formed by that triangle will be zero, that is no such triangle can be formed. We can use the area of triangle formula and apply collinearity condition there to do the proof.
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