MHT CET · Maths · Pair of Lines
If the lines represented by \(\left(\mathrm{k}^2+2\right) \mathrm{x}^2+3 \mathrm{xy}-6 \mathrm{y}^2=0\) are perpendicular to each other, then the values of \(\mathrm{K}\) are
- A \(\pm 3\)
- B \(\pm 4\)
- C \(\pm 1\)
- D \(\pm 2\)
Answer & Solution
Correct Answer
(D) \(\pm 2\)
Step-by-step Solution
Detailed explanation
The lines \(\left(k^2+2\right) x^2+3 x y-6 y^2=0\) are perpendicular to each other.
\(\therefore\left(\mathrm{k}^2+2\right)+(-6)=0 \Rightarrow \mathrm{k}^2=4 \Rightarrow \mathrm{k}= \pm 2\)
\(\therefore\left(\mathrm{k}^2+2\right)+(-6)=0 \Rightarrow \mathrm{k}^2=4 \Rightarrow \mathrm{k}= \pm 2\)
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