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Question
If the pair of straight lines x2 – 2kxy – y2 = 0 bisect the angle between the pair of straight lines x2 – 2lxy – y2 = 0, Show that the later pair also bisects the angle between the former
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Solution
The equations of the given pair of straight lines are
x2 – 2kxy – y2 = 0 .......(1)
x2 – 2lxy – y2 = 0 .......(2)
Given that the pair x2 – 2kxy – y2 = 0 bisects the angle between the pair x2 – 2lxy – y2 = 0
∴ The equation of the bisector of the pair
x2 – 2lxy – y2 = 0 is the pair x2 – 2kxy – y2 = 0
The equation of the bisector of x2 – 2lxy – y2 = 0 is
`(x^2 - y^2)/(1 - ( - 1)) = (xy)/(- l)`
`(x^2 - y^2)/2 = (xy)/(- l)`
`x^2 - y^2 = - 2/l xy`
`x^2+2/l xy-y^2 =0` .....(3)
Equation (3) and Equation (1) represents the same straight lines.
∴ The coefficients are proportional.
`1/1 = (2/l)/(- 2"k") = (- 1)/(- 1)`
1 = `1/((l"k")` = 1
`l"k"` = – 1 .....(4)
To show that the pair x2 – 2lxy – y2 = 0 bisects the angle between the pair x2 – 2kxy – y2 = 0, it is enough to prove the equation of the bisector of x2 – 2kxy – y2 = 0 is x2 – 2lxy – y2 = 0
The equation of the bisector of x2 – 2kxy – y2 = 0 is
`(x^2 - y^2)/(1 - ( - 1)) = (xy)/(- "k")`
`(x^2 - y^2)/2 = (xy)/(- "k")`
`x^2 - y^2 = 2/"k" xy`
`x^2 - y^2 = - 2/(1/l) xy`
By equationn (4)
x2 – y2 = 2lxy .
x2 – 2lxy – y2 = 0
∴ The pair x2 – 2lxy – y2 = 0 bisects the angle between the pair x2 – 2kxy – y2 = 0
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