Question

Find the value of $k$ if $\left( k+1,2k \right),\left( 3k,2k+3 \right),\left( 5k-1,5k \right)$ are collinear.

Answer 1

First check whether the given points are distinct or not by equating respective coordinates of a pair of points. If they are distinct, equate the determinant from the formula of the area of the triangle to zero and put the points. $$$$

Complete step by step answer:
We call more than two points co-linear when all of the points lie on the line joined by first two points.$$$$
We know that if $\left( {{x}_{1}},{{y}_{1}} \right)$,$\left( {{x}_{1}},{{y}_{1}} \right)$,$\left( {{x}_{1}},{{y}_{1}} \right)$ are three distinct points in the Cartesian plane , then the area of triangle made by these three points is given by the modulus of determinant $D=\dfrac{1}{2}\left| \begin{matrix} 1 & {{x}_{1}} & {{y}_{1}} \\\ 1 & {{x}_{2}} & {{y}_{2}} \\\ 1 & {{x}_{3}} & {{y}_{3}} \\\ \end{matrix} \right|$. They are going to be collinear if and only if the value of the determinant is zero. In symbols,

1 & {{x}_{1}} & {{y}_{1}} \\\ 1 & {{x}_{2}} & {{y}_{2}} \\\ 1 & {{x}_{3}} & {{y}_{3}} \\\ \end{matrix} \right|=0$$ We expand by the second column and use the fact that area is positive quantity to take modulus wherever necessary. We have $$D={{x}_{1}}\left( {{y}_{2}}-{{y}_{1}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right)=0...(1)$$ The given points are $\left( k+1,2k \right)$,$\left( 3k,2k+3 \right)$,$\left( 5k-1,5k \right)$. We are asked to find the value of $k$. Lets us denote them as A$\left( k+1,2k \right)$, B $\left( 3k,2k+3 \right)$ ,C$\left( 5k-1,5k \right)$ respectively.$$$$ Case-1:$$$$ The three points are going to collinear if they are the same point not distinct points that means $A=B=C$. Let us take $A=B$. We equate respective first and second coordinates of A and B and get, $$ \begin{aligned} & \left( k+1,2k \right)=\left( 3k,2k+3 \right) \\\ & \Rightarrow k+1=3k,2k=2k+3 \\\ & \Rightarrow k=\dfrac{1}{2},k\in R \\\ \end{aligned}$$ So the only value of $k$ for $A=B$ is $\dfrac{1}{2}$. Again we take $B=C$ and equate respective first and second coordinates of B and C. We have, $$ \begin{aligned} & ~\left( 3k,2k+3 \right)=\left( 5k-1,5k \right) \\\ & \Rightarrow 5k-1=3k,2k+3=5k \\\ & \Rightarrow k=\dfrac{1}{2},k=1 \\\ \end{aligned} $$ $k$ cannot have two different values. $B\ne C$ for any $k$. Hence the points are distinct.$$$$ Case-2:$$$$ We now know the points are distinct. Let us use the area of triangle formula for the points$A\left( k+1,2k \right)=A\left( {{x}_{1}},{{y}_{1}} \right)$, $B\left( 3k,2k+3 \right)=B\left( {{x}_{2}},{{y}_{2}} \right)$ and $C\left( 5k-1,5k \right)=C\left( {{x}_{3}},{{y}_{3}} \right)$ to check whether they are collinear or not. We have $$\begin{aligned} & D=\left( k+1 \right)\left( 3-3k \right)+3k\left( 3k \right)+\left( 5k-1 \right)\left( -{{3}_{2}} \right)=0 \\\ & \Rightarrow 6{{k}^{2}}-5k+6=0 \\\ & \\\ \end{aligned}$$ The above equation is a quadratic equation whose roots are the value of $k$. We solve it by splitting the middle term and get $$\begin{aligned} & \Rightarrow 6{{k}^{2}}-5k+6=0 \\\ & \Rightarrow 6{{k}^{2}}-12k-3k+6 \\\ & \Rightarrow 6k\left( k-2 \right)-3\left( k-2 \right)=0 \\\ & \Rightarrow \left( k-2 \right)\left( 6k-3 \right)=0 \\\ \end{aligned}$$ **So the value $k=2$ or $k=\dfrac{3}{6}=\dfrac{1}{2}$ $$$$** **Note:** The key in this problem is the modulus we take for area because area is positive. You can verify whether the points are collinear by putting the value of $k$. You can also solve this problem using a section formula where a point divides a line segment with a certain ratio.