Question

Find the value of a for which the following points A (a, 3), B (2, 1) and C (5, a) are collinear. Hence, find the equation of the line.

Answer 1

Hint: Assume the given points as A \[\left( {{x}_{1}},{{y}_{1}} \right)\], B \[\left( {{x}_{2}},{{y}_{2}} \right)\] and C \[\left( {{x}_{3}},{{y}_{3}} \right)\] and apply the formula for area of triangle given as: - Area = \[\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]\]. Substitute this area expression equal to 0 and find the values of a. Now, for each value of calculate the equation of line by using the formula: - \[\left( y-{{y}_{1}} \right)=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right)\]. Here, we have been provided with three points A (a, 3), B (2, 1) and C (5, a) and we have been asked to determine the value of ‘a’ for which these points will be collinear.

Complete step by step answer:
Now, let us assume the given points as A \[\left( {{x}_{1}},{{y}_{1}} \right)\], B \[\left( {{x}_{2}},{{y}_{2}} \right)\] and C \[\left( {{x}_{3}},{{y}_{3}} \right)\]. So, we have,
\[\begin{align}
  & \Rightarrow \left( {{x}_{1}},{{y}_{1}} \right)=\left( a,3 \right) \\
 & \Rightarrow \left( {{x}_{2}},{{y}_{2}} \right)=\left( 2,1 \right) \\
 & \Rightarrow \left( {{x}_{3}},{{y}_{3}} \right)=\left( 5,a \right) \\
\end{align}\]
We know that collinear points means that all the given points lie on a single straight line and the triangle formed by three collinear points has an area equal to 0. So, applying the formula for area of a triangle in coordinate geometry, we get,
\[\Rightarrow \] Area (\[\Delta ABC\]) = \[\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]\]
Substituting the values of coordinates and Area = 0, we get,
\[\begin{align}
  & \Rightarrow 0=\dfrac{1}{2}\left[ a\left( 1-a \right)+2\left( a-3 \right)+5\left( 3-1 \right) \right] \\
 & \Rightarrow 0=\left[ a-{{a}^{2}}+2a-6+10 \right] \\
 & \Rightarrow 0=-{{a}^{2}}+3a+4 \\
 & \Rightarrow {{a}^{2}}-3a-4=0 \\
\end{align}\]
Using the middle term split method, we have,
\[\begin{align}
  & \Rightarrow {{a}^{2}}-4a+a-4=0 \\
 & \Rightarrow a\left( a-4 \right)+1\left( a-4 \right)=0 \\
 & \Rightarrow \left( a+1 \right)\left( a-4 \right)=0 \\
\end{align}\]
\[\Rightarrow \left( a+1 \right)=0\] or \[\left( a-4 \right)=0\]
\[\Rightarrow a=-1\] or \[a=4\]
Therefore, there can be two possible values of a for which the given points A, B and C are collinear.
Now, we have to find the equation of the line.
We know that equation of a straight line is given as, \[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\], where ‘m’ is the slope of the line given by the formula: - \[\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)=m\]. Let us select two points A and B to determine the slopes of the line. So, let us consider two values of ‘a’ one – by – one.
1. Case (1): - When a = -1.
For a = -1, we have the points A and B as (-1, 3) and (2, 1) respectively. So, we have,
\[\begin{align}
  & \Rightarrow m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \\
 & \Rightarrow m=\dfrac{1-3}{2-\left( -1 \right)} \\
 & \Rightarrow m=\dfrac{-2}{3} \\
\end{align}\]
Applying the formula for equation of line, we get,
\[\begin{align}
  & \Rightarrow \left( y-3 \right)=\dfrac{-2}{3}\left( x-\left( -1 \right) \right) \\
 & \Rightarrow \left( y-3 \right)=\dfrac{-2}{3}\left( x+1 \right) \\
\end{align}\]
By cross – multiplication, we have,
\[\begin{align}
  & \Rightarrow 3\left( y-3 \right)=-2\left( x+1 \right) \\
 & \Rightarrow 3y-9=-2x-2 \\
 & \Rightarrow 2x+3y-7=0 \\
\end{align}\]
2. Case (2): - When a = 4.
For a = 4, we have the points A and B as (4, 3) and (2, 1) respectively. So, we have,
\[\begin{align}
  & \Rightarrow m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \\
 & \Rightarrow m=\dfrac{1-3}{2-4} \\
 & \Rightarrow m=\dfrac{-2}{-2} \\
 & \Rightarrow m=1 \\
\end{align}\]
Applying the formula for equation of line, we get,
\[\begin{align}
  & \Rightarrow \left( y-3 \right)=1\left( x-\left( -1 \right) \right) \\
 & \Rightarrow y-3=x+1 \\
 & \Rightarrow x-y+4=0 \\
\end{align}\]

Note:
 One must remember the definition of collinear points and its condition to solve the question. You may see that we have considered both the values of ‘a’ to find the equation of the line. This is because both the values are valid and cannot be rejected. In the above solution, we have considered points A and B to find the slopes and equations of the line. You may also select points B and C or A and C for the same results.