Question

an tricky question plz solve it​

Answer 1

using area equal to zero.solve it

Answer 2

$\underline{\underline{\Huge{\textsf{Question:}}}}$

$\sf\textsf{If the Points A(K+1, 2K), B(3K, 2K+3) and C(5K-1, 5K)}\\\sf\textsf{are collinear then find value of K.}$





$\underline{\Huge{\textsf{Concept Behind:}}}$

$\maltese$ The three Points $\mathsf{A(x_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3}, y_{3})}$ are said to be $\textsf{collinear}$, if they $\textit{don't form a triangle}$ when then these points are joined.

In other words,

$\maltese\textbf{For Collinear Points, the Area of triangle}\\\textbf{formed by joining three points will be 0.}$

$\maltese$ Area of Triangle formed by the Points $\mathsf{A(x_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3}, y_{3})}$ is given by:

$\quad \LARGE{\boxed{\mathbf{\triangle= \dfrac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})}}}$





$\underline{\underline{\Huge{\textsf{Solution:}}}}$

$\underline{\LARGE{\textsf{Step-1:}}}$
$\sf\textsf{Note the given Points}$

Here,

$A = (x_{1}, y_{1}) = (K+1, 2K),\\B = (x_{2}, y_{2}) = (3K, 2K+3)\\C = (x_{3}, y_{3}) = (5K-1, 5K)$




$\underline{\LARGE{\textsf{Step-2:}}}$

$\sf\textsf{Substitute in [1] and Equate Area to zero}\\\sf\textsf{to obtain Quadratic equation}$

$\begin{aligned}\implies \frac{1}{2}\: [ (K+1)(2K+3-5K)+3K(5K-2K)+(5K-1)(2K-2K-3)] &= 0\\\\\implies \frac{1}{2}\: [ (K+1)(3-3K)+3K(3K)+(5K-1)(-3)]&= 0\\\\\implies \frac{1}{2}\: [(K+1)(3-3K)+3K(3K)+(5K-1)(-3)]&= 0\\\\\implies 3K-3K^{2}+3-3K+9x^{2}-15 K+3 &= 0\\\\\implies -3K^{2}+9K^{2}-15k+6&=0\\\\\implies 6k^{2}-15K+6& =0\\\\\implies3(2K^{2}-5K+2)&=0\\\\\implies2K^{2}-5K+2 &=0\end{aligned}$




$\underline{\LARGE{\textsf{Step-3:}}}$

$\sf\textsf{Factorise the Quadratic equation}\\\sf\textsf{to obtain value of K}$

$\begin{array}{crclc}&&4K^2&&\\&&\swarrow\searrow&&\\&-4K&&-K&\end{array}\quad\boxed{\begin{minipage}{6cm}\item \bigstar\quad\; \, 2K^{2}-5K+2=0 \\\begin{itemize}\item 2K^2\times 2\qquad\; \, = 4t^2 \item 4K^2\qquad\qquad\, = (-4K)\times (-K) \item (-4K)+(-K) \quad= -5K\end{itemize}\end{minipage}}$

$\begin{aligned}\implies \sf 2K^{2}-5K+2&=0\\\\\implies \sf 2K^{2}-4x-K+2&=0\\\\\implies \sf 2K(K-2)-1(K-2)&=0\\\\\implies \sf (2K-1)(K-2)&=0\\\\\implies 2K-1 = 0 \: &\: and &K-2 = 0\\\\\implies \bf K= \frac{1}{2} \: &\: and& \bf K = 2\end{aligned}$





$\therefore$

$\blacksquare\; \textsf{The value of K for which the points Points A(K+1, 2K), B(3K, 2K+3) and C(5K-1, 5K)}\\\sf\textsf{will be Collinear is } \Large{\underline{\mathbf{\frac{1}{2}\: or\: 2}}}$