Question

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

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

Answer 1

Answer: (C)

$a=1+b$

If the points $(a,b),(b,a)$ and $(a^2,{-b}^2)$ are collinear then

$\Delta =\left|\begin{array}{ccc}a& b& 1\\ b& a& 1\\ a^2& -b^2& 1\end{array}\right|=0R_1=R_1-R_2\Rightarrow \Delta =\left|\begin{array}{ccc}a-b& b-a& 0\\ b& a& 1\\ a^2& -b^2& 1\end{array}\right|=0R_2=R_2-R_3\Rightarrow \Delta =\left|\begin{array}{ccc}a-b& b-a& 0\\ b-a^2& a+b^2& 0\\ a^2& -b^2& 1\end{array}\right|=0$

Expanding along $C_3$

$\Rightarrow (a-b)(a+b^2)-(b-a)(b-a^2)=0\Rightarrow (a-b)(a+b^2+b-a^2)=0\Rightarrow a+b^2+b-a^2=0\Rightarrow b^2-a^2=-b-a\Rightarrow (b-a)(b+a)=-(b+a)\Rightarrow b-a=-1\Rightarrow a=1+b$

So, option C is correct.