Answer:
Solution given:
The given equation of a line is
ax²+2hxy+by²=0
Let y=mx be any one line of
ax²+2hxy+by²=0
Let the perpendicular line of
y=mx is
x+my=0
According to the question the line x+my=0 is one line of Ax²+2Hxy +By²=0
Substituting value of y
Ax²+2Hx [tex] \frac{ - x}{m} [/tex]+[tex]B \frac{x²}{m²} [/tex]=0
Ax²-[tex] \frac{ 2H}{m}x² [/tex]+[tex]B \frac{x²}{m²} [/tex]=0
Taking LCM
we get
Ax²m²-2mHx²+Bx²=0
x²[Am²-2Hm+B]=0..............[1]
Again.
ax²+2hx(mx)+B(mx)²=0
ax²+2hmx²+Bm²x²=0
x²[a+2hm+Bm²]=0
bm²+2hn+a=0.....................[2]
Taking coefficient of equation 1 &2.
equation 1. b 2h a b 2h
equation 2.A. -2h B A -2H
Doing criss cross multiplication
ignore first coefficient and repeat first and second
again
lines are perpendicular so
[tex] \frac{m²}{2hB} [/tex]=[tex] \frac{m}{aA-bB} [/tex]=[tex]\frac{1}{-2Hb-2hA} [/tex]
Taking 1st & 2nd ratio,we get,
m=[tex] \frac{2hB+2Ha}{aA-bB} [/tex]....[*]
Taking 3rd & 2nd ratio,we get,
m=[tex] \frac{aA-bB}{-2Hb-2hA} [/tex] ....[#]
Again
Equating equation * &# we get;
[tex] \frac{2hB+2Ha}{aA-bB} [/tex]=[tex] \frac{aA-bB}{-2Hb-2hA} [/tex]
(aA-bB)²=-4(hB+Ha)(Hb+HA)
(aA-bB)²+4(hB+Ha)(Hb+HA)=0 is a required condition.
