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Question: Which of the following points are collinear? A). \((2a, 0), (3a, 0), (a, 2a)\) B). \((3a, 0), (0...

Which of the following points are collinear?
A). (2a,0),(3a,0),(a,2a)(2a, 0), (3a, 0), (a, 2a)
B). (3a,0),(0,3b),(a,2b)(3a, 0), (0, 3b), (a, 2b)
C). (3a,b),(a,2b),(−a,b)(3a, b), (a, 2b), (-a, b)
D). (a,−6),(−a,3b),(−2a,−2b)(a, -6), (-a, 3b), (-2a, -2b)

Explanation

Solution

In order to prove the collinearity of points, we need to ensure that they all lie on the straight line. So, if we find an equation of the line joining the first two points and then the third point satisfies this equation, we can declare that the three points are collinear. The equation of the line joining two points is,
y−y1x−x1=y2−y1x2−x1\dfrac{y-{{y}_{1}}}{x-{{x}_{1}}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}

Complete step-by-step solution:
By the collinearity of some points, we mean that all the points lie on the same straight line. In other words, we can say that if we can draw a common straight line through all the points, then those points are said to be collinear. This means that if we are given a straight line, then all the points lying on it are collinear to one another.
In order to prove the collinearity of points, we need to ensure that they all lie on the straight line. So, if we find an equation of the line joining the first two points and then the third point satisfies this equation, we can declare that the three points are collinear. The equation of the line joining two points is,
y−y1x−x1=y2−y1x2−x1\dfrac{y-{{y}_{1}}}{x-{{x}_{1}}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}
For the first option, the equation of the line joining (2a,0),(3a,0)\left( 2a,0 \right),\left( 3a,0 \right) is,
y−0x−2a=0−03a−2a ⇒y=0 \begin{aligned} & \dfrac{y-0}{x-2a}=\dfrac{0-0}{3a-2a} \\\ & \Rightarrow y=0 \\\ \end{aligned}
Putting the third point, we get,
⇒2a=0\Rightarrow 2a=0 which is not true. So, the first option is incorrect.
For the second option, the equation of the line joining (3a,0),(0,3b)\left( 3a,0 \right),\left( 0,3b \right) is,
y−0x−3a=3b−00−3a ⇒y=−ba(x−3a) \begin{aligned} & \dfrac{y-0}{x-3a}=\dfrac{3b-0}{0-3a} \\\ & \Rightarrow y=-\dfrac{b}{a}\left( x-3a \right) \\\ \end{aligned}
Putting the third point, we get,
⇒2b=−ba(a−3a)=2b\Rightarrow 2b=-\dfrac{b}{a}\left( a-3a \right)=2b which is true. So, the second option is correct.
For the third option, the equation of the line joining (3a,b),(a,2b)\left( 3a,b \right),\left( a,2b \right) is,
y−bx−3a=2b−ba−3a ⇒y=b−b2a(x−3a) \begin{aligned} & \dfrac{y-b}{x-3a}=\dfrac{2b-b}{a-3a} \\\ & \Rightarrow y=b-\dfrac{b}{2a}\left( x-3a \right) \\\ \end{aligned}
Putting the third point, we get,
⇒b=b−b2a(−a−3a)=3b\Rightarrow b=b-\dfrac{b}{2a}\left( -a-3a \right)=3b which is not true. So, the third option is incorrect.
For the fourth option, the equation of the line joining (a,−b),(−a,3b)\left( a,-b \right),\left( -a,3b \right) is,
y+bx−a=3b+b−a−a ⇒y=−b−2ba(x−a) \begin{aligned} & \dfrac{y+b}{x-a}=\dfrac{3b+b}{-a-a} \\\ & \Rightarrow y=-b-\dfrac{2b}{a}\left( x-a \right) \\\ \end{aligned}
Putting the third point, we get,
⇒−2b=b−b2a(−2a−a)=52b\Rightarrow -2b=b-\dfrac{b}{2a}\left( -2a-a \right)=\dfrac{5}{2}b which is not true. So, the fourth option is incorrect.
Thus, we can conclude that the correct option is option B.

Note: We can also solve the problem in another way. If the slopes of the line joining the first two points and the line joining the last two points are the same, then the three points are collinear. The slope of the line joining two points is y2−y1x2−x1\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} . For the first option, (0−03a−2a=0)≠(2a−0a−3a=−1)\left( \dfrac{0-0}{3a-2a}=0 \right)\ne \left( \dfrac{2a-0}{a-3a}=-1 \right) . For the second, (3b−00−3a=−ba)=(2b−3ba−0=−ba)\left( \dfrac{3b-0}{0-3a}=-\dfrac{b}{a} \right)=\left( \dfrac{2b-3b}{a-0}=-\dfrac{b}{a} \right) . For the third, (2b−ba−3a=−b2a)≠(b−2b−a−a=b2a)\left( \dfrac{2b-b}{a-3a}=-\dfrac{b}{2a} \right)\ne \left( \dfrac{b-2b}{-a-a}=\dfrac{b}{2a} \right) . For the fourth option, (3b+b−a−a=−2ba)≠(−2b−3b−2a+a=−5ba)\left( \dfrac{3b+b}{-a-a}=-\dfrac{2b}{a} \right)\ne \left( \dfrac{-2b-3b}{-2a+a}=-\dfrac{5b}{a} \right) .