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The points $(-a,-b),(a, b),(0,0)$ and $\left(a^{2}, a b\right), a \neq 0, b \neq 0$ are always
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Verified Answer
The correct answer is:
collinear
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.
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.
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