Question

The nature of straight lines represented by the equation $4x^2+12xy+9y^2=0,$ is

• A. Real and coincident
• B. Real and different
• C. Imaginary and different
• D. None of the above

Answer 1

The correct option is A

Real and coincident

Explanation for the correct option:

Given: $4x^2+12xy+9y^2=0$

We know that for a pair of straight lines given by $a{x}^2+2hxy+b{y}^2=0,$

If $h^2–ab>0$, then the lines are real and distinct.

If $h^2–ab=0$, then the lines are real and coincident.

If $h^2–ab<0$, then the lines are imaginary.

So, for the given equation, $4x^2+12xy+9y^2=0$

We have $a=4,b=9$ and $h=6$.

Now,

$$\begin{gathered} h^2–ab={\left(6\right)}^2-\left(4\right)\left(9\right) \\ =36-36 \\ =0 \end{gathered}$$

Therefore, the lines are real and coincident.

Hence, option A is the correct .