aai 2018

Question

 

For what values of k, the points (k,2-2k),(-k+1,2k) and (-4-k,6-2k) are collinear?

• A.  ½  ,-½  
• B.  -1 ,½ 
• C.  0, 1
• D.  -1 , 1

Comments

Yes,$2$ is the correct answer.

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@pream sagar

can you tell me what is the problem you are facing while typing the question instead of posting screen shot ?

please edit all your own questions yourself.... make them text, if really screen shot is needed then there is no problem !

Answer 1

If given points  three points $(x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3})$ are colinear then area of triangle should be $0$

      Area of triangle $=\frac{1}{2}\begin{vmatrix} x_{1}-x_{2}&x_{2}-x_{3} \\ y_{1}-y_{2}&y_{2}-y_{3} \end{vmatrix}=0$

Here $(x_{1},y_{1})=(k,2-2k),(x_{2},y_{2})=(-k+1,2k),(x_{3},y_{3})=(-4-k,6-2k)$

              $\frac{1}{2}\begin{vmatrix} 2k-1&5 \\2-4k&4k-6\end{vmatrix}=0$

                $\begin{vmatrix} 2k-1&5 \\2-4k&4k-6\end{vmatrix}=0$

          $\Rightarrow(2k-1)(4k-6)-5(2-4k)=0$

        $\Rightarrow(2k-1)(4k-6)+5.2(2k-1)=0$

        $\Rightarrow(2k-1)[4k-6+10]=0$

        $\Rightarrow(2k-1)(4k+4)=0$

       $\Rightarrow 2k-1=0$  $(or)$  $4k+4=0$

      $\Rightarrow 2k=1$  $(or)$  $4k=-4$

    $\Rightarrow k=\frac{1}{2}$  $(or)$  $k=-1$

Comments

@Mk Utkarsh

please verify my procedure is right?

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Lakshman any reference of above method, I know one method in which C3 is 1.

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see this please

https://www.urbanpro.com/gre/how-to-determine-if-points-are-collinear

https://www.quora.com/How-do-I-prove-that-three-points-are-collinear