Question

If the points A (k + 1, 2k), B (3k, 2k + 3), and C (5k - 1, 5k) are collinear then find the value of k.

Answer 1

Hint: As given in the question that the given points are collinear. So, for collinear points we know that the area of the triangle formed by them is zero. By applying the area of the triangle of coordinate geometry we get the value of k.
Complete step-by-step answer:
A ($x_1$, $y_1$), B ($x_2$, $y_2$) and C ($x_3$, $y_3$) are all three vertices of the triangle ABC.
Now, the area of triangle formula used when coordinates are given:
Area of $\Delta ABC$ $=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$
Given: A (k + 1, 2k), B (3k, 2k + 3), and C (5k - 1, 5k) are the collinear points. So, the area of the triangle formed by them is zero.
Area of triangle $=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$
As we know that if the points are collinear, then the area of the triangle is zero. On putting the values of coordinates, we get
$\begin{align}
  & \Rightarrow 0=\dfrac{1}{2}\left[ \left( k+1 \right)\left( 2k+3-5k \right)+3k\left( 5k-2k \right)+\left( 5k-1 \right)\left( 2k-2k-3 \right) \right] \\
 & \Rightarrow 0=\left( k+1 \right)\left( 3-3k \right)+3k\left( 3k \right)+\left( 5k-1 \right)\left( -3 \right) \\
 & \Rightarrow 0=3k-3{{k}^{2}}+3-3k+9{{k}^{2}}-15k+3 \\
 & \Rightarrow 0=6{{k}^{2}}-15k+6 \\
\end{align}$
Taking 3 common from right hand side, we have:
$\Rightarrow 0=3\left( 2{{k}^{2}}-5k+2 \right)$
Shifting 3 in left hand side from right hand side, the term will become zero,
$\Rightarrow 0=2{{k}^{2}}-5k+2$
Factorisation using middle term splitting of the above equation, we get
$\begin{align}
  & \Rightarrow 0=2{{k}^{2}}-4k-k+2 \\
 & \Rightarrow 0=2k\left( k-2 \right)-1\left( k-2 \right) \\
 & \Rightarrow 0=\left( 2k-1 \right)\left( k-2 \right) \\
\end{align}$
So, one of them may be zero or both will be zero. Using this we get all the values of k.
$\begin{align}
  & \Rightarrow 2k-1=0\text{ or }k-2=0 \\
 & \Rightarrow k=\dfrac{1}{2}\text{ or }k=2 \\
\end{align}$
Hence, the values of k are $\dfrac{1}{2},2$.
Note: This problem can be alternatively solved by using the section formula which can be stated as: $\left( x,y \right)=\left( \dfrac{m{{x}_{1}}+{{x}_{2}}}{m+1},\dfrac{m{{y}_{1}}+1{{y}_{2}}}{m+1} \right)\text{ in ratio 1}:m$. Now, let B divide A and C in the ratio 1: m. By putting the values in the above formula, we obtain two equations corresponding to x and y coordinates. So, the values of K and m are evaluated.