Question

the pair of lines 9x2+y2+6xy-0=0 are
1) parallel and not coincident
2) imaginary
3) perpendicular
4) none of the above

Answer 1

Answer:

1

Step-by-step explanation:

Given equation

9x^2+y^2+6xy-4

(3x)^2+y^2+2*3xy=4

(3x+y)^2=4

(3x+y)=+/-4

3x+y=2

3x+y=-2

So, constant term is different

Therefore they are parallel and not coincident

Answer 2

Answer:

The given lines are parallel and not coincident.

Step-by-step explanation:

Given the pair of lines,  $9x^2+y^2+6xy-0=0$

correcting the equation to $9x^2+y^2+6xy-4=0$

Rewriting the polynomial,  $(3x)^2+2(3x)y+y^2=4$

in the form of algebraic identity,  $a^2+b^2+2ab=(a+b)^{2}$

we get, $(3x+y)^2=4$

Taking square root on both sides, $\sqrt{(3x+y)^2} =\sqrt{4}$  

$\implies 3x+y=\pm4$

Hence, the separate equations of the lines are  $3x+y=4~\&~3x+y=-4$

In a pair of linear equations,  $a_1x+b_1y=c_1~\&~a_2x+b_2y=c_2$

• if $a_1b_2=a_2b_1$, then they represent parallel lines and

• if $a_1a_2+b_1b_2=0$, then they represent perpendicular lines

From the pair of equations we have,

$3\times 1=3\times 1$ , therefore, the equations represent parallel lines.

Since $c_1 ~\&~c_2$  are different, the equations are not coincident lines.

Therefore, the given lines are parallel and not coincident.

So, the correct answer is option 1.