Question

Find the value of $k$ for which the points $A\left( {k + 1,2k} \right),B(3k,2k + 3),C(5k - 1,5k)$ are collinear.

Answer 1

Hint:We are to find the value of $k$ for which the three points are collinear.For this, first we must know what collinear points are. So, collinear points are those which lie on the same line. For example, if $A,B$ and $C$ are said to be collinear points if they lie on the same line. So, if three points $A,B,C$ are collinear points then the area of the triangle formed by the three points will be zero, as a line doesn’t occupy any area.

Complete step by step answer:
The given points are:
$A(k + 1,2k),B(3k,2k + 3),C(5k - 1,5k)$
Now, for the points to be collinear, they must satisfy the condition;
Area $\Delta ABC = 0$
Let us assume $ABC$ is a triangle with vertices $A({x_1},{y_1}),B({x_2},{y_2}),C({x_3},{y_3})$.
Therefore, A($\Delta ABC$)$ = \dfrac{1}{2}[{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})]$
$\Rightarrow \dfrac{1}{2}[(k + 1)\{ (2k + 3) - (5k)\} + (3k)\{ (5k) - (2k)\} + (5k - 1)\{ (2k) - (2k + 3)\} ]$
\[\Rightarrow \dfrac{1}{2}[(k + 1)(3 - 3k) + (3k)(3k) + (5k - 1)( - 3)]\]
Simplifying the expression we get,
$\Rightarrow \dfrac{1}{2}[( - 3{k^2} + 3) + (9{k^2}) + ( - 15k + 3)]$
$\Rightarrow \dfrac{1}{2}[6{k^2} - 15k + 6]$
$\Rightarrow \dfrac{3}{2}[2{k^2} - 5k + 2]$
Now, Area of triangle $\Delta ABC$$ = 0$
$ \Rightarrow \dfrac{3}{2}[2{k^2} - 5k + 2] = 0$
$ \Rightarrow 2{k^2} - 5k + 2 = 0$
Splitting the middle term we get,
$ \Rightarrow 2{k^2} - 4k - k + 2 = 0$
$ \Rightarrow 2k(k - 2) - 1(k - 2) = 0$
$ \Rightarrow (2k - 1)(k - 2) = 0$
$ \therefore k = \dfrac{1}{2},2$

Therefore, the value of $k$ for which $A,B,C$ are collinear are $\dfrac{1}{2},2$.

Note:Another way of solving these kinds of problems is by using the concept of distance between two points. For example, if $A,B$ and $C$ are collinear points, that is they lie on the same line. So, if three points $A,B,C$ are collinear and $B$ lies in between $A$ and $C$, then, the distance between $A$ and $B$ plus the distance between $B$ and $C$ is equal to the distance between $A$ and $C$. That is AB + BC = AC$.