Question

If the equation \(x^2-ky^2-4x+6y-5 = 0\) represents a pair of straight lines, then their point of intersection is

• A. \((1,2)\)
• B. \((2,3)\)
• C. \((3,4)\)
• D. \((4,5)\)

Hint:

Find k from the pair-of-lines determinant condition, then solve the two partial derivatives equal to zero.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A general second degree equation represents a pair of lines when its determinant \(\begin{vmatrix}a&h&g\\h&b&f\\g&f&c\end{vmatrix}\) is zero. The point of intersection satisfies \(\frac{\partial S}{\partial x}=0\) and \(\frac{\partial S}{\partial y}=0\).

Step 2: Find k:
Here \(a=1\), \(b=-k\), \(c=-5\), \(h=0\), \(g=-2\), \(f=3\). The condition \(abc+2fgh-af^2-bg^2-ch^2=0\) gives
\[ 5k-9+4k=0 \Rightarrow k=1 \]

Step 3: Find the intersection:
\(\frac{\partial S}{\partial x}=2x-4=0\) gives \(x=2\). \(\frac{\partial S}{\partial y}=-2ky+6=0\) gives \(y=\frac{3}{k}=3\).

Step 4: Check:
Put \(k=1\): \(x^2-y^2-4x+6y-5=(x-2)^2-(y-3)^2=0\). This is the pair \(x-y+1=0\) and \(x+y-5=0\), which meet at \((2,3)\). The other options do not make both factors vanish.

Final Answer:
The lines meet at \((2,3)\), option (B). \[ \boxed{(2,3)} \]