Let the equation of the pair of lines, y = px and y = qx, can be written as (y $$-$$ px) (y $$-$$ qx) = 0. Then the equation of the pair of the angle bisectors of the lines x2 $$-$$ 4xy $$-$$ 5y2 = 0 is :
JEE · Math · previous-year question
- A.x2 $$-$$ 3xy + y2 = 0
- B.x2 + 4xy $$-$$ y2 = 0
- C.x2 + 3xy $$-$$ y2 = 0correct
- D.x2 $$-$$ 3xy $$-$$ y2 = 0
Answer
C. x2 + 3xy $$-$$ y2 = 0
Explanation
Equation of angle bisector of homogeneous equation of pair of straight line ax2 + 2hxy + by2 is $${{{x^2} - {y^2}} \over {a - b}} = {{xy} \over h}$$ for x2 – 4xy – 5y2 = 0 a = 1, h = – 2, b = – 5 So, equation of angle bisector is $${{{x^2} - {y^2}} \over {1 - ( - 5)}} = {{xy} \over { - 2}}$$ $${{{x^2} - {y^2}} \over 6} = {{xy} \over { - 2}}$$ $$ \Rightarrow {x^2} - {y^2} = - 3xy$$ So, combined equation of angle bisector is $$ {x^2} + 3xy - {y^2} = 0$$
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