Q71 / 155 · WBJEE 2016 · Mathematics

The points $(-a,-b),(a, b),(0,0)$ and $\left(a^{2}, a b\right), a \neq 0, b \neq 0$ are always

  • Correct answer
    A
    collinear
  • B
    vertices of a parallelogram
  • C
    vertices of a rectangle
  • D
    lie on a circle

Solution

Let the four points be $A(-a,-b), B(a, b), C(0,0)$ and $D\left(a^{2}, a b\right)$ If $A,B$ and $C$ are collinear. Then, $\left|\begin{array}{ccc}-a & -b & 1 \\ a & b & 1 \\ 0 & 0 & 1\end{array}\right|=0$ $\Rightarrow \quad-a(b-0)+b(a-0)+1(0)=0$ $-a b+a b=0$ Hence, $A, B$ and $C$ ae collinear. Again, if $B, C$ and $D$ are collinear. Then, $\left|\begin{array}{ccc}a & b & 1 \\ 0 & 0 & 1 \\ a^{2} & a b & 1\end{array}\right|=0$ $\Rightarrow \quad a(0-a b)-b\left(0-a^{2}\right)+1(0) ;=0$ $-a^{2} b+a^{2} b=0$ $\Rightarrow$ Hence, $B, C$ and $D$ are collinear. So, the points $A, B, C$ and $D$ are akways collinear.

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

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