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If one of the two lines 3x^2 +kxy-y^2=0 bisects an angle between the co-ordinate axes, then k=?​

Answer»

Step-by-step EXPLANATION:

\sf  {3x}^{2}  - kxy -  {y}^{2}  = 0 \\  \\  \\  \sf divided \:  \: by \:  \:  {x}^{2}  \\  \\  \\  \sf 3 -  \frac{ky}{x}  -  \frac{ {y}^{2} }{ {x}^{2} }  = 0 \\  \\  \\  \sf y = mx \\  \\  \\  \sf  \frac{y}{x}  = m \\  \\  \\  \sf  {m}^{2}  + km - 3 = 0 \\  \\  \\  \sf \: one \:  \: of \: \:  the \:  \: line \:  \: bisect \:  \: co -  drdinate \:  \: axis \\  \\  \\  \sf put \:  \: m = 1 \:  \: or \:  \: m =  - 1 \\  \\  \\  \sf 1 + k - 3 = 0 \\  \\  \\  \sf k = 2  \\  \\  \\  \sf 1 - k - 3 = 0 \\  \\  \\  \sf k =  - 2 \\  \\  \\  \sf \fbox{ \sf k = ±2 }



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