Question

Show that the equation,<br /><br />${x}^{2} - 3xy + 2 {y}^{2} - 2x - 3y - 35 = 0$<br /><br />for every real value of 'x' there is a real value of 'y' , and for every real value of 'y' there is a real value of 'x'.<br /><br /><br />❌❌NO SPAMMING❌❌​

Answer 1

Answer:

MATHS

The pairs of straight lines x

2

−3xy+2y

2

=0 and x

2

−3xy+2y

2

+x−2=0 form a

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ANSWER

Given pairs

x

2

−3xy+2y

2

=0

(x−2y)(x−y)=0

x−2y=0 and x−y=0 are two sides

x

2

−3xy+2y

2

+x−2=0

x=

2

−(1−3y)±

1+9y

2

−6y−4(2y

2

−2)

x=

2

−(1−3y)±

1+9y

2

−6y−8y

2

+8)

2x=−1+3y±

y

2

−6y+9)

2x=−1+3y±

(y−3)

2

)

2x=−1+3y±(y−3)

2x=−1+3y+y−3 and 2x=−1+3y−y+3

2x−4y=−4 and 2x−2y=2

x−2y=−2 and x−y=1 are other two sides

x−2y=0 and x−y=0 and x−2y=−2 and x−y=1 are four sides

Here two opposite sides are parallel

Angle between x−2y=0 and x−y=0 by formula is

tanθ=

∣

∣

∣

∣

∣

1+m

1

m

2

m

1

−m

2

∣

∣

∣

∣

∣

tanθ=

∣

∣

∣

∣

∣

∣

∣

∣

1+

2

1

1−

2

1

∣

∣

∣

∣

∣

∣

∣

∣

tanθ=

3

1

Here the sides are not perpendicular

Hence it is parallelogram