Question

If the lines $a₁x+b₁y+c₁=0$, $a₁x+b₁y+c₂=0$, $a₂x+b₂y+d₁=0$ and $a₂x+b₂y+d₂=0$ are sides of a rhombus, then

• A. $(a_{2}^{2}+b_{2}^{2})(c_{1}-c_{2})^{2}=(a_{1}^{2}+b_{1}^{2})(d_{1}-d_{2})^{2}$
• B. $(a_{1}^{2}+b_{1}^{2})|d_{1}-d_{2}|=(a_{2}^{2}+b_{2}^{2})|c_{1}-c_{2}|$
• C. $(a_{2}^{2}+b_{2}^{2})(d_{1}-d_{2})^{2}=(a_{1}^{2}+b_{1}^{2})(c_{1}-c_{2})^{2}$
• D. $(a_{1}^{2}+b_{1}^{2})|c_{1}-c_{2}|=(a_{2}^{2}+b_{2}^{2})|d_{1}-d_{2}|$

Hint:

If the lines $a1x+b1y+c1=0$, $a1x+b1y+c2=0$, $a2x+b2y+d1=0$ and $a2x+b2y+d2=0$ are sides of a rhombus, then

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
In a rhombus, the perpendicular distance between each pair of parallel sides must be equal.
Step 2: Evaluation
Distance between first pair: $h_{1} = \frac{|c_{1}-c_{2}|}{\sqrt{a_{1}^{2}+b_{1}^{2}}}$.
Distance between second pair: $h_{2} = \frac{|d_{1}-d_{2}|}{\sqrt{a_{2}^{2}+b_{2}^{2}}}$.
Step 3: Final Calculation
Setting $h_{1} = h_{2}$: $\frac{|c_{1}-c_{2}|}{\sqrt{a_{1}^{2}+b_{1}^{2}}} = \frac{|d_{1}-d_{2}|}{\sqrt{a_{2}^{2}+b_{2}^{2}}}$.
Step 4: Conclusion
Squaring and cross-multiplying gives $(a_{2}^{2}+b_{2}^{2})(c_{1}-c_{2})^{2} = (a_{1}^{2}+b_{1}^{2})(d_{1}-d_{2})^{2}$.
Hence, the Answer is: (a)