MathematicsMedium128×since 2002Q5504The equation of one of the straight lines which passes through the point (1, 3) and makes an angles tan−1(2){\tan ^{ - 1}}\left( {\sqrt 2 } \right)tan−1(2) with the straight line, y + 1 = 32{\sqrt 2 }2 x is :A42x+5y−(15+42)=04\sqrt 2 x + 5y - \left( {15 + 4\sqrt 2 } \right) = 042x+5y−(15+42)=0B52x+4y−(15+42)=05\sqrt 2 x + 4y - \left( {15 + 4\sqrt 2 } \right) = 052x+4y−(15+42)=0C42x+5y−42=04\sqrt 2 x + 5y - 4\sqrt 2 = 042x+5y−42=0D42x−5y−(5+42)=04\sqrt 2 x - 5y - \left( {5 + 4\sqrt 2 } \right) = 042x−5y−(5+42)=0Check answerSkip