IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry - Pair of st. lines-equation of angle bisectors

rohitkuruvila

look,ive found a way of doing it. take the first 3 terms of the equation

ax² + 2hxy + by² + 2gx + 2fy + c = 0

ie,ax² + 2hxy + by²=0(these lines have the the same slope as the previous lines,only their point of intersection is the origin)

the org. equation is 3x² + 8xy - 3y² - 20x - 10y + 25 = 0

the new eq.is 3x² + 8xy - 3y²=0

the equation of their angle bisectors is (x²- y²)/xy=(a - b)/h

after putting the values ,you get 2x² - 3xy - 2y²=0

now just shift this equation to the org.point of intersection(2,1)

you get

2(x - 2)² - 3(x - 2)(y - 1) - 2(y - 1)=0

expand this eq. and you'll get the eq. of the angle bisectors.

nudge me if im wrong.