Question

For any real value of $\lambda$ the equation $2{{x}^{2}}+3{{y}^{2}}-8x-6y+11-\lambda =0$ doesn’t represents a pair of straight lines?
A. True
B. False

Answer 1

In this question we have to find that the given statement is true or false. We will analyze the necessary condition for the equation to represent a pair of straight lines. We will try to factorize the LHS of the given equation and get the desired answer.

Complete step by step solution:
We have been given that for any real value of $\lambda$ the equation $2{{x}^{2}}+3{{y}^{2}}-8x-6y+11-\lambda =0$ doesn’t represent a pair of straight lines.
We have to find whether the given statement is true or false.
Now, we know that if the equation $a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0$represents a pair of straight line then $a,b,h\ne 0$ and ${{h}^{2}}-ab\ge 0$.
Now, comparing the general equation with given equation we will get the values
$\Rightarrow a=2,b=3,h=0,g=-4,f=-3,c=11$
Now, the condition $a,b,h\ne 0$ is not satisfied as we have $h=0$.
Let us check the second condition ${{h}^{2}}-ab\ge 0$.
Now, substituting the values we will get
$\begin{aligned} & \Rightarrow {{0}^{2}}-6\ge 0 \\\ & \Rightarrow -6\ge 0 \\\ \end{aligned}$
Also the second condition is not satisfied as we get the negative value.
So for any value of $\lambda$ the equation $2{{x}^{2}}+3{{y}^{2}}-8x-6y+11-\lambda =0$ doesn’t represents a pair of straight lines.
Hence the given statement is true.

So, the correct answer is “Option A”.

Note: The point to be noted is that if the equation represents a pair of straight lines then the left side of the equation must be resolved into two linear factors. The second degree terms cannot be factored into two linear factors so the equation doesn’t represent a pair of straight lines.