Question

The equation \(x^2 - Ky^2 - 4x + 6y - 5 = 0\) represents a pair of straight lines. Find the point of intersection.

Hint:

Try treating the equation as a quadratic in one variable, for instance $x$. For it to represent two real straight lines, the term under the square root of the quadratic formula must turn out to be a perfect square; this directly gives the unknown constant and lets you factor the equation.

Answer 1

Step 1: Understanding the Concept:
The point of intersection of a pair of straight lines \(S(x,y) = 0\) is the point where the partial derivatives \(\frac{\partial S}{\partial x} = 0\) and \(\frac{\partial S}{\partial y} = 0\) vanish.

Step 2: Key Formula or Approach:
1. Condition for pair of lines: \(\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0\).
2. Point of intersection: Solve \(\frac{\partial S}{\partial x} = 0\) and \(\frac{\partial S}{\partial y} = 0\).

Step 3: Detailed Explanation:
Let \(S(x,y) = x^2 - Ky^2 - 4x + 6y - 5 = 0\).
First, find \(K\) using the condition \(\Delta = 0\):
\(a=1, b=-K, c=-5, g=-2, f=3, h=0\).
\[ 1(-K)(-5) + 0 - 1(3)^2 - (-K)(-2)^2 - 0 = 0 \] \[ 5K - 9 + 4K = 0 \implies 9K = 9 \implies K = 1 \] Now find the intersection point using partial derivatives:
\(\frac{\partial S}{\partial x} = 2x - 4 = 0 \implies x = 2\).
\(\frac{\partial S}{\partial y} = -2Ky + 6 = 0\).
Substituting \(K=1\):
\(-2(1)y + 6 = 0 \implies 2y = 6 \implies y = 3\).
The point is \((2,3)\).

Step 4: Final Answer:
The point of intersection is \((2, 3)\).

Answer 2

Concept:

• If a second-degree equation represents a pair of straight lines, solving it as a quadratic in one variable must give an expression that is linear in the other variable, which means the term under the square root of the quadratic formula must be a perfect square trinomial.
• Once the unknown constant is found this way, factoring directly gives the equations of the two individual lines, and their point of intersection is simply the solution of those two linear equations solved together.

Step 1: Treat the given equation as a quadratic in $x$.
$x^2 - Ky^2 - 4x + 6y - 5 = 0$
$x^2 - 4x - (Ky^2 - 6y + 5) = 0$
$x = \dfrac{4 \pm \sqrt{16 + 4(Ky^2 - 6y + 5)}}{2} = 2 \pm \sqrt{Ky^2 - 6y + 9}$

Step 2: Force the expression under the square root to be a perfect square trinomial in $y$.
$Ky^2 - 6y + 9$ should match $(ay+b)^2 = a^2y^2 + 2aby + b^2$
Constant term: $b^2 = 9 \Rightarrow b = 3$
Middle term: $2ab = -6 \Rightarrow a = -1$
Leading term: $a^2 = K \Rightarrow K = 1$

Step 3: Substitute $K = 1$ and factor the square root term.
$Ky^2 - 6y + 9 = y^2 - 6y + 9 = (y-3)^2$
$x = 2 \pm (y - 3)$
Line 1: $x = y - 1$, i.e. $x - y + 1 = 0$
Line 2: $x = 5 - y$, i.e. $x + y - 5 = 0$

Step 4: Solve the two lines together to find their point of intersection.
$x - y + 1 = 0$ ... (i)
$x + y - 5 = 0$ ... (ii)
Adding (i) and (ii): $2x - 4 = 0 \Rightarrow x = 2$
Substituting into (ii): $2 + y - 5 = 0 \Rightarrow y = 3$

Final Answer: The point of intersection is $(2, 3)$