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JulijaS [17]
3 years ago
15

Give me a complete and managed answer.​

Mathematics
2 answers:
Sauron [17]3 years ago
8 0

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.

nikdorinn [45]3 years ago
6 0

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

<u>According</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>question</u><u> </u><u>the</u><u> </u><u>line</u><u> </u><u>x</u><u>+</u><u>my</u><u>=</u><u>0</u><u> </u><u>is</u><u> </u><u>one</u><u> </u><u>line</u><u> </u><u>of</u><u> </u><u>Ax²</u><u>+</u><u>2</u><u>H</u><u>x</u><u>y</u><u> </u><u>+</u><u>By²</u><u>=</u><u>0</u>

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<u> </u><u>i</u><u>s</u><u> </u><u>a</u><u> </u><u>r</u><u>e</u><u>q</u><u>u</u><u>i</u><u>r</u><u>e</u><u>d</u><u> </u><u>c</u><u>o</u><u>n</u><u>d</u><u>i</u><u>t</u><u>i</u><u>o</u><u>n</u><u>.</u>

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