Question

If the point $(2,-3)$ lies on $k{{x}^{2}}-3{{y}^{2}}+2x+y-2=0$, then $k$ is equal to
A. $\dfrac{1}{7}$
B. $16$
C. $7$
D. $12$

Answer 1

Hint: In this question, we are to find the value of $k$. Since the given point lies on the given equation of pair of straight lines, by substituting the point into the given equation, we get the required $k$ value.



Formula Used:The combined equation of pair of straight lines is written as
$H\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}=0$
This is called a homogenous equation of second degree in $x$ and $y$
And
$S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
This is called a general equation of second degree in $x$ and $y$.
If ${{h}^{2}}If ${{h}^{2}}=ab$, then $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents coincident lines.
If ${{h}^{2}}>ab$, then $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents two real and different lines that pass through the origin.
Thus, the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents two lines. They are:
$ax+hy\pm y\sqrt{{{h}^{2}}-ab}=0$



Complete step by step solution:Given equation of the pair of lines is
$k{{x}^{2}}-3{{y}^{2}}+2x+y-2=0\text{ }...(1)$
The point $(2,-3)$ lie on the given lines.
Since the point lies on the pair of straight lines, the point satisfies the given equation.
So, by substituting, the given point in (1), we get
$\begin{align}
  & k{{x}^{2}}-3{{y}^{2}}+2x+y-2=0 \\
 & \Rightarrow k{{(2)}^{2}}-3{{(-3)}^{2}}+2(2)+(-3)-2=0 \\
 & \Rightarrow 4k-27+4-5=0 \\
 & \Rightarrow 4k=28 \\
 & \, therefore, k=7 \\
\end{align}$
Therefore, the value of the required variable is $k=7$.



Option ‘C’ is correct



Note: Here, we need to remember that, any point that lies on a line will satisfy the equation of the line. So, by this, since the given point lies on the given combined equation of the pair of lines, the point satisfies the given equation. So, on substituting, we get the required variable value.