Question

Find the value of a for which the following points $A\left( a,3 \right)$, $B\left( 2,1 \right)$ and $C\left( 5,a \right)$ are collinear. Also find the equation of the line passing through these points?

Answer 1

We start solving the problem by using the property that if three points are collinear, then the area of the triangle formed by those three points as vertices is zero. We then apply area of the triangle formed by three points as vertices $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ is $\left| \begin{matrix} {{x}_{1}}-{{x}_{2}} & {{x}_{1}}-{{x}_{3}} \\\ {{y}_{1}}-{{y}_{2}} & {{y}_{1}}-{{y}_{3}} \\\ \end{matrix} \right|$ to find the value of a. We then find the points using the value of a and then find the equation of the line using the fact that the equation of the line passing through the points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ is $y-{{y}_{1}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\times \left( x-{{x}_{1}} \right)$.

Complete step by step answer:
According to the problem, we are asked to find the value of ‘a’ if the points $A\left( a,3 \right)$, $B\left( 2,1 \right)$ and $C\left( 5,a \right)$ are collinear. We then find the equation of the line passing through these points.

We know that if three points are collinear, then the area of the triangle formed by those three points as vertices is zero.
We know that the area of the triangle formed by three points as vertices $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ is $\left| \begin{matrix} {{x}_{1}}-{{x}_{2}} & {{x}_{1}}-{{x}_{3}} \\\ {{y}_{1}}-{{y}_{2}} & {{y}_{1}}-{{y}_{3}} \\\ \end{matrix} \right|$.
So, we have given that the area of the triangle formed by the points $A\left( a,3 \right)$, $B\left( 2,1 \right)$ and $C\left( 5,a \right)$ is 0.
$\Rightarrow \left| \begin{matrix} a-2 & a-5 \\\ 3-1 & 3-a \\\ \end{matrix} \right|=0$.
$\Rightarrow \left| \begin{matrix} a-2 & a-5 \\\ 2 & 3-a \\\ \end{matrix} \right|=0$.
We know that $\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=\left( a\times d \right)-\left( b\times c \right)$.
$\Rightarrow \left( \left( a-2 \right)\times \left( 3-a \right) \right)-\left( \left( a-5 \right)\times 2 \right)=0$.
$\Rightarrow \left( 3a-{{a}^{2}}-6+2a \right)-\left( 2a-10 \right)=0$.
$\Rightarrow -{{a}^{2}}-6+5a-2a+10=0$.
$\Rightarrow -{{a}^{2}}+3a+4=0$.
$\Rightarrow -{{a}^{2}}+4a-a+4=0$.
$\Rightarrow -a\left( a-4 \right)-1\left( a-4 \right)=0$.
$\Rightarrow \left( -a-1 \right)\left( a-4 \right)=0$.
$\Rightarrow -a-1=0$, $a-4=0$.
$\Rightarrow a=-1$, $a=4$.
∴ The values of ‘a’ are –1, 4.
Now, let us find the equation of the line when $a=-1$.
So, we have the line passing through the points $A\left( -1,3 \right)$, $B\left( 2,1 \right)$. We know that the equation of the line passing through the points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ is $y-{{y}_{1}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\times \left( x-{{x}_{1}} \right)$.
$\Rightarrow y-3=\left( \dfrac{1-3}{2+1} \right)\times \left( x+1 \right)$.
$\Rightarrow y-3=\left( \dfrac{-2}{3} \right)\times \left( x+1 \right)$.
$\Rightarrow 3y-9=-2x-2$.
$\Rightarrow 2x+3y-7=0$.
∴ The equation of the line is $2x+3y-7=0$.
Now, let us find the equation of the line when $a=4$.
So, we have the line passing through the points $A\left( 4,3 \right)$, $B\left( 2,1 \right)$.
$\Rightarrow y-3=\left( \dfrac{1-3}{2-4} \right)\times \left( x+1 \right)$.
$\Rightarrow y-3=\left( \dfrac{-2}{-2} \right)\times \left( x+1 \right)$.
$\Rightarrow y-3=1\times \left( x+1 \right)$
$\Rightarrow 3y-9=x+1$.
$\Rightarrow x-3y+10=0$.

∴ The equation of the line is $x-3y+10=0$.

Note: We can also find the values of ‘a’ as shown below:
We know that the points are said to be collinear if they all lie on same line. Since the points A, B and C lie on the same line, we know that the slope of AB = slope of BC.
We know that the slope of the line passing through the points $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ is $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
$\Rightarrow \dfrac{1-3}{2-a}=\dfrac{a-1}{5-2}$.
$\Rightarrow \dfrac{-2}{2-a}=\dfrac{a-1}{3}$.
$\Rightarrow -2\times 3=\left( a-1 \right)\times \left( 2-a \right)$.
$\Rightarrow -6=2a-{{a}^{2}}-2+a$.
$\Rightarrow -6=-{{a}^{2}}+3a-2$.
$\Rightarrow {{a}^{2}}-3a-4=0$.
$\Rightarrow {{a}^{2}}-4a+a-4=0$.
$\Rightarrow a\left( a-4 \right)+1\left( a-4 \right)=0$.
$\Rightarrow \left( a+1 \right)\left( a-4 \right)=0$.
$\Rightarrow a+1=0$, $a-4=0$.
$\Rightarrow a=-1$, $a=4$.