A) a circle
B) two lines
C) two parallel lines
D) two mutually perpendicular lines
Correct Answer: D
Solution :
\[{{(x+y)}^{2}}-({{x}^{2}}+{{y}^{2}})=0\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}+2xy-{{x}^{2}}-{{y}^{2}}=0\] \[\Rightarrow \] \[2xy=0\] On comparing with \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] Here, \[a=0,b=0,h=1\] \[\therefore \] \[\tan \theta =\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}=\infty \] \[\Rightarrow \] \[\theta ={{90}^{o}}\] \[\therefore \] The given equation represents two mutually perpendicular lines. Note: In second degree homogeneous equation if\[\text{a + b = 0,}\] then the pair of lines are perpendicular.
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