WBJEE · Maths · Straight Lines
The points \((-a,-b),(a, b),(0,0)\) and \(\left(a^{2}, a b\right), a \neq 0, b \neq 0\) are always
- A collinear
- B vertices of a parallelogram
- C vertices of a rectangle
- D lie on a circle
Answer & Solution
Correct Answer
(A) collinear
Step-by-step Solution
Detailed explanation
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,…
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