Q119 / 170 · Manipal (MET) 2015 · Mathematics

One possible condition for the three points $(a, b),(b, a)$ and $\left(a^2,-b^2\right)$ to be collinear, is

  • A
    $a-b=2$
  • B
    $a+b=2$
  • Correct answer
    C
    $a=1+b$
  • D
    $a=1-b$

Solution

Given points will be collinear, if $\left|\begin{array}{ccc}a & b & 1 \\ b & a & 1 \\ a^2 & -b^2 & 1\end{array}\right|=0$ $\Rightarrow \quad\left|\begin{array}{ccc}a & b & 1 \\ b-a & a-b & 0 \\ a^2-a & -b^2-b & 0\end{array}\right|=0$ [applying $R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R_1$ ] $\begin{array}{lr}\Rightarrow & (a-b)\left|\begin{array}{ccc}a & b & 1 \\ -1 & 1 & 0 \\ a^2-a & -b^2-b & 0\end{array}\right|=0 \\ \Rightarrow & (a-b)\left(b^2+b-a^2+a\right)=0\end{array}$ $\begin{array}{rr}\Rightarrow & (a-b)\left\{(a+b)-\left(a^2-b^2\right)\right\}=0 \\ \Rightarrow & (a-b)(a+b)(1-a+b)=0 \\ \Rightarrow & a=b \text { or } a+b=0 \text { or } a=1+b\end{array}$

Hence, the correct answer is option: 'C'.

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