show that the equation of the pair of straight lines passing through the origin and making an angle 30 with the line 3x-y-1=0 is 13x^-12xy-3y^
Answers
Given : equation of the pair of straight lines passing through origin and making an angle of 30° with the line 3x-y-1=0 is 13 x² - 12xy -3y²=0
To Find : Show that
Solution:
Let say slope of line making an angle of 30° with the line 3x-y-1=0 is m
slope of line 3x-y-1=0 is 3 as ( y = 3x - 1)
slope m & 3
tan 30° = | (m - 3) /( 1 + 3m)|
±1/√3 = (m - 3) /( 1 + 3m)
=> 1 + 3m = m√3 - 3√3 or 1 + 3m = - m√3 + 3√3
1 + 3m = m√3 - 3√3
=> m(3 - √3) = - (3√3 + 1)
=> m = - (3√3 + 1) /((3 - √3)
=> m = - (3√3 + 1)(3 + √3)/6
=> m = - (12 + 10√3 )/6
=> m = - (6 + 5√3)/3
y = - (6 + 5√3)x/3 as passes though origin
=> 3y + (6 + 5√3)x = 0
1 + 3m = - m√3 + 3√3
=> m(3 + √3) = 3√3 - 1
=> m = (3√3 - 1)(3 - √3) / 6
=> m = (-12 + 10√3)/6
=> m = -( 6 - 5√3)/3
=> =3y + ( 6 - 5√3)x 0
3y + (6 + 5√3)x = 0
3y + ( 6 - 5√3)x 0
Multiplying both
9y² + 3y(12)x - 39x² = 0
=> 3y² + 12xy - 13x² = 0
=> 13x² - 12xy + 3y² = 0
QED
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