Question

To the lines $ax²+2hxy+by²=0$ the lines $a²x²+2h(a+b)xy+b²y²=0,$ are

• A. equally inclined
• B. perpendicular
• C. bisector of the angle
• D. None of the above

Hint:

To the lines $ax=0$ the lines $ax+2h(a+b)xy+by=0,$ are

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Two pairs of lines are equally inclined if they share the same angle bisectors.
Step 2: Evaluation
Bisectors of $ax^{2}+2hxy+by^{2}=0$ are $\frac{x^{2}-y^{2}}{a-b} = \frac{xy}{h}$.
Step 3: Final Calculation
Bisectors of $a^{2}x^{2}+2h(a+b)xy+b^{2}y^{2}=0$ are $\frac{x^{2}-y^{2}}{a^{2}-b^{2}} = \frac{xy}{h(a+b)}$. Dividing both sides by $(a+b)$ gives $\frac{x^{2}-y^{2}}{a-b} = \frac{xy}{h}$.
Step 4: Conclusion
Since the bisector equations are identical , the lines are equally inclined.
Hence, the Answer is: (a)