Question

Prove that the following equation represent pair of parallel lines by equation $9x^2-6xy+y^2+18x-6y+8=0$.

Answer 1

Given equation is $9x^2-6xy+y^2+18x-6y+8=0$

Comparing with $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

$a=9,h=-3,b=1,g=9,f=-3,c=8$

Consider, $abc+2fgh-a{f}^2-b{g}^2-c{h}^2$

$\Rightarrow$ $9\times 1\times 8+(-6)\times 9\times 3-9\times 9-1\times 81-8\times 9$

$\Rightarrow$ $72+162-81-81-72$

$\Rightarrow$ $0$

$\therefore$ The given equation represents a pair of lines ---- ( 1 )

Now, $h^2-ab=(-3{)}^2-9\times 1=9-9=0$

As $h^2-ab=0$ lines are parallel. ------ ( 2 )

$\therefore$ From ( 1 ) and ( 2 )

$9x^2-6xy+y^2+18x-6y+8=0$ represents a pair of parallel straight lines.