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 line.

Answer 1

Hint: The points A, B, c are collinear. So the slope of AB is equal to the slope of BC. Find the slope of AB and BC. Equate them together and simplify it, to get the equation of line.

Complete step-by-step answer:
Collinear points are the points that lie on the same line. If two or more than two points lie on a line close to or far from each other, then they are said to be collinear. We can find collinear points using the slope formula.
The slope formula is used when there are 3 or more points and are collinear if the slope of any 2 pairs of points is the same.
Here we have been given three points A (a, 3), B (2, 1) and C (5, a).
These points are collinear. So, we can say that,
Slope of AB = Slope of BC = Slope of AC, as the point A, B and C is collinear.
Equation to find slope = \[\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
Here, \[A\left( a,3 \right)=\left( {{x}_{1}},{{y}_{1}} \right)\]
\[\begin{align}
  & B\left( 2,1 \right)=\left( {{x}_{2}},{{y}_{2}} \right) \\
 & C\left( 5,a \right)=\left( {{x}_{3}},{{y}_{3}} \right) \\
\end{align}\]
\[\therefore \] Slope of AB = Slope of BC can be taken,
\[\begin{align}
  & \therefore \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{{{y}_{3}}-{{y}_{2}}}{{{x}_{3}}-{{x}_{2}}} \\
 & \dfrac{1-3}{2-a}=\dfrac{a-1}{5-2}\Rightarrow \dfrac{-2}{2-a}=\dfrac{a-1}{3} \\
 & -6=\left( 2-a \right)\left( a-1 \right) \\
 & -6=2a-2-{{a}^{2}}+a \\
\end{align}\]
\[\Rightarrow -6=3a-2-{{a}^{2}}\], thus remaining the equation we get,
\[\therefore {{a}^{2}}-3a-6+2=0\]
\[{{a}^{2}}-3a-4=0\], hence we got the equation of line.
The above equation is in the form of a quadratic equation, \[a{{x}^{2}}+bx+c=0\]. Thus comparing we get,
a = 1, b = -3, c = -4
Putting these values in \[a=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
\[\begin{align}
  & \therefore a=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 1\times \left( -4 \right)}}{2\times 1}=\dfrac{3\pm \sqrt{9+16}}{2}=\dfrac{3\pm \sqrt{25}}{2} \\
 & \therefore a=\dfrac{3\pm 5}{2} \\
\end{align}\]
\[a=\dfrac{3+5}{2}=\dfrac{8}{2}\] and \[a=\dfrac{3-5}{2}=\dfrac{-2}{2}\]
\[a=4\] and \[a=-1\]
Thus when a = 4, the points are A (4, 3), B (2, 1) and C (5, 4).
When a = -1, the points are A (-1, 3), B (2, 1) and C (5, -1).
Hence we got the equation of line as \[{{a}^{2}}-3a-4=0\] and the values of a = 4, -1.

Note: We can also find an equation using the area of the triangle.
Area of triangle \[=\dfrac{1}{2}\left| \begin{matrix}
   {{x}_{1}}-{{x}_{2}} & {{x}_{2}}-{{x}_{3}} \\
   {{y}_{1}}-{{y}_{2}} & {{y}_{2}}-{{y}_{3}} \\
\end{matrix} \right|=0\]
                              \[=\dfrac{1}{2}\left| \begin{matrix}
   a-2 & 2-5 \\
   3-1 & 1-a \\
\end{matrix} \right|=0\]
\[\begin{align}
  & \dfrac{1}{2}\left\{ \left[ \left( a-2 \right)\left( 1-a \right) \right]-\left[ \left( 2-5 \right)\left( 3-1 \right) \right] \right\}=0 \\
 & \left( a-{{a}^{2}}-2+2a \right)-\left( -3\times 2 \right)=0 \\
 & -{{a}^{2}}+3a-2+6=0 \\
 & -{{a}^{2}}+3a+4=0 \\
 & \therefore {{a}^{2}}-3a-4=0 \\
\end{align}\]