Question:
The locus of the equation $xy+yz=0$ is
The locus of the equation $xy+yz=0$ is
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The locus of the equation $xy+yz=0$ is
Updated On: Apr 15, 2026
- a pair of straight lines
- a pair of parallel lines
- a pair of perpendicular planes
- None of the above
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Factorize the equation $xy+yz=0 \Rightarrow y(x+z)=0$.
Step 2: Analysis
This gives two planes: $y = 0$ and $x + z = 0$.
Step 3: Evaluation
Normal to $y=0$ is $(0, 1, 0)$. Normal to $x+z=0$ is $(1, 0, 1)$.
Step 4: Conclusion
Dot product of normals: $(0)(1) + (1)(0) + (0)(1) = 0$. The planes are perpendicular.
Final Answer: (c)
Factorize the equation $xy+yz=0 \Rightarrow y(x+z)=0$.
Step 2: Analysis
This gives two planes: $y = 0$ and $x + z = 0$.
Step 3: Evaluation
Normal to $y=0$ is $(0, 1, 0)$. Normal to $x+z=0$ is $(1, 0, 1)$.
Step 4: Conclusion
Dot product of normals: $(0)(1) + (1)(0) + (0)(1) = 0$. The planes are perpendicular.
Final Answer: (c)
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