Question

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

Answer 1

Hint: Collinear points are the points that lie on the same line. Two or more than two points lie on a line close to or far from each other.


Complete step by step solution:

Points $A(k + 1,2k),B(3k,2k + 3)andC(3k - 1,5k)$ are collinear, then the area of triangle must be $0$
$\therefore $Area of triangle $ABC = 0$
Here, ${x_1} = k + 1,\,{x_2} = 3k,\,{x_3} = 3k - 1$
${y_1} = 2k,\,{y_2} = 2k + 3,\,{y_3} = 5k$
Area of triangle $ABC = \dfrac{1}{2}|({x_1}({y_1} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})|$
$ = \dfrac{1}{2}|(k + 1)\{ (2k + 3) - 5k\} + 3k\{ 5k - 2k\} + (5k - 1)\{ 2k - (2k + 3)\} |$
$ = \dfrac{1}{2}|(k + 1)\{ 2k + 3 - 5k\} + 3k(3k) + (5k - 1)\{ 2k - 2k - 3\} |$
$ = \dfrac{1}{2}|(k + 1)\{ 3 - 3k\} + 3k(3k) + (5k - 1)\{ - 3\} |$
\[ = \dfrac{1}{2}|\{ k(3 - 3k) + 1(3 - 3k)\} + \{ 3 \times 3 \times k \times k\} + \{ - 15k + 3\} |\]
$ = \dfrac{1}{2}|\{ 3k - 3{k^2} + 3 - 3k\} + \{ 9{k^2}\} - 15k + 3|$
$ = \dfrac{1}{2}|\{ 3k - 3{k^2} - 3k + 3\} + 9{k^2} - 15k + 3|$
$ = \dfrac{1}{2}| - 3{k^2} + 3 + 9{k^2} - 15k + 3|$
$ = \dfrac{1}{2}(6{k^2} - 15k + 6)$
$ = \dfrac{1}{2}(6{k^2} - 15k + 6)$
$\therefore $Area of $\Delta ABC = 0$ $[\therefore po\operatorname{int} s\,\,are\,\,collinear]$
$0 = \dfrac{1}{2}(6{k^2} - 15k + 6)$
$0 = 6{k^2} - 15k + 6$
$0 = 3(2{k^2} - 5k + 2)$
$\dfrac{0}{3} = 2{k^2} - 5k + 2$
$0 = 2{k^2} - 5k + 2$
$ \Rightarrow 2{k^2} - 5k + 2 = 0$
We will split the middle term in the above equation
$2{k^2} - 5k + 2 = 0$
$2{k^2} - 4k - k + 2 = 0$
$2k(k - 2) - 1(k - 2) = 0$
$(2k - 1)(12 - 2) = 0$
$
   \Rightarrow 2k - 1 = 0 \\
   \Rightarrow 2k = 0 + 1 \\
   \Rightarrow 2k = 1 \\
   \Rightarrow K = \dfrac{1}{2} \\
 $
$
  k - 2 = 0 \\
  k = 2 \\
 $
$ \Rightarrow 12 = 2,\dfrac{1}{2}$
The points of $k$ are $2\,and\,\dfrac{1}{2}$.


Note: We can find the collinearity of the points by using the method mentioned below:



 $A,B,C$ are collinear points if they satisfy the condition. $AC = AB + BC$