A) \[{{a}^{2}}+{{d}^{2}}=2ac\]
B) \[{{a}^{2}}+{{d}^{2}}=2bd\]
C) \[{{a}^{2}}+ac+bd+{{d}^{2}}=0\]
D) \[{{a}^{2}}+{{d}^{2}}=4bc\]
Correct Answer: C
Solution :
The given equation being homogeneous of third degree represents three straight lines through the origin. Since two of these lines are to be at right angles. Let pair of these lines be \[({{x}^{2}}+pxy-{{y}^{2}})\], p is constant and other factor is \[(ax-dy)\]. Hence,\[a{{x}^{3}}+b{{x}^{2}}y+cx{{y}^{2}}+d{{y}^{3}}=({{x}^{2}}+pxy-{{y}^{2}})(ax-dy)\] Comparing the coefficients of similar terms, we get \[b=ap-d\] .....(i); \[c=-pd-a\] .....(ii) Multiplying (i) by d and (ii) by a and adding, we get \[bd+ac=-{{d}^{2}}-{{a}^{2}}\ \Rightarrow \ {{a}^{2}}+ac+bd+{{d}^{2}}=0\].
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