Question

Give me a complete and managed answer.​

Answer 1

If one of lines given by the first pair by y=mx, then by the given condition one of the lines given by the 2nd pair should be

y = −(1/3)x.

∴ bm² + 2hm + a = 0.

and b(− 1/m)² + 2h(− 1/m) + d = 0 or dm²− 2hm + b = 0.

Ans. (ad − bb)² − 4(hb + ha)(bh + hd) = 0.

Answer 2

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 $\frac{ - x}{m}$+$B \frac{x²}{m²}$=0

Ax²-$\frac{ 2H}{m}x²$+$B \frac{x²}{m²}$=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

$\frac{m²}{2hB}$=$\frac{m}{aA-bB}$=$\frac{1}{-2Hb-2hA}$

Taking 1st & 2nd ratio,we get,

m=$\frac{2hB+2Ha}{aA-bB}$....[*]

Taking 3rd & 2nd ratio,we get,

m=$\frac{aA-bB}{-2Hb-2hA}$ ....[#]

Again

Equating equation * &# we get;

$\frac{2hB+2Ha}{aA-bB}$=$\frac{aA-bB}{-2Hb-2hA}$

(aA-bB)²=-4(hB+Ha)(Hb+HA)

(aA-bB)²+4(hB+Ha)(Hb+HA)=0 is a required condition.